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10.8: Use Radicals in Functions


Learning Objectives

By the end of this section, you will be able to:

  • Evaluate a radical function
  • Find the domain of a radical function
  • Graph radical functions

Before you get started, take this readiness quiz.

  1. Solve: (1−2x≥0).
    If you missed this problem, review Example 2.50.
  2. For (f(x)=3x−4), evaluate (f(2),f(−1),f(0)).
    If you missed this problem, review Example 3.48.
  3. Graph (f(x)=sqrt{x}). State the domain and range of the function in interval notation.
    If you missed this problem, review Example 3.56.

Evaluate a Radical Function

In this section we will extend our previous work with functions to include radicals. If a function is defined by a radical expression, we call it a radical function.

  • The square root function is (f(x)=sqrt{x}).
  • The cube root function is (f(x)=sqrt[3]{x}).

Definition (PageIndex{1}): radical function

A radical function is a function that is defined by a radical expression.

To evaluate a radical function, we find the value of (f(x)) for a given value of (x) just as we did in our previous work with functions.

Example (PageIndex{1})

For the function (f(x)=sqrt{2 x-1}), find

  1. (f(5))
  2. (f(-2))

Solution:

a.

(f(x)=sqrt{2 x-1})

To evaluate (f(5)), substitute (5) for (x).

(f(5)=sqrt{2 cdot 5-1})

Simplify.

(f(5)=sqrt{9})

Take the square root.

(f(5)=3)

b.

(f(x)=sqrt{2 x-1})

To evaluate (f(-2)), substitute (-2) for (x).

(f(-2)=sqrt{2(-2)-1})

Simplify.

(f(-2)=sqrt{-5})

Since the square root of a negative number is not a real number, the function does not have a value at (x=-2).

Exercise (PageIndex{1})

For the function (f(x)=sqrt{3 x-2}), find

  1. (f(6))
  2. (f(0))
Answer
  1. (f(6)=4)
  2. no value at (x=0)

Exercise (PageIndex{2})

For the function (g(x)=sqrt{5x+5}), find

  1. (g(4))
  2. (g(-3))
Answer
  1. (g(4)=5)
  2. no value at (f(-3))

We follow the same procedure to evaluate cube roots.

Example (PageIndex{2})

For the function (g(x)=sqrt[3]{x-6}), find

  1. (g(14))
  2. (g(-2))

Solution:

a.

(g(x)=sqrt[3]{x-6})

To evaluate (g(14)), substitute (14) for (x).

(g(14)=sqrt[3]{14-6})

Simplify.

(g(14)=sqrt[3]{8})

Take the cube root.

(g(14)=2)

b.

(g(x)=sqrt[3]{x-6})

To evaluate (g(-2)), substitute (-2) for (x).

(g(-2)=sqrt[3]{-2-6})

Simplify.

(g(-2)=sqrt[3]{-8})

Take the cube root.

(g(-2)=-2)

Exercise (PageIndex{3})

For the function (g(x)=sqrt[3]{3 x-4}), find

  1. (g(4))
  2. (g(1))
Answer
  1. (g(4)=2)
  2. (g(1)=-1)

Exercise (PageIndex{4})

For the function (h(x)=sqrt[3]{5 x-2}), find

  1. (h(2))
  2. (h(-5))
Answer
  1. (h(2)=2)
  2. (h(-5)=-3)

The next example has fourth roots.

Example (PageIndex{3})

For the function (f(x)=sqrt[4]{5 x-4}), find

  1. (f(4))
  2. (f(-12))

Solution:

a.

(f(x)=sqrt[4]{5 x-4})

To evaluate (f(4)), substitute (4) for (x).

(f(4)=sqrt[4]{5 cdot 4-4})

Simplify.

(f(4)=sqrt[4]{16})

Take the fourth root.

(f(4)=2)

b.

(f(x)=sqrt[4]{5 x-4})

To evaluate (f(-12)), substitute (-12) for (x).

(f(-12)=sqrt[4]{5(-12)-4})

Simplify.

(f(-12)=sqrt[4]{-64})

Since the fourth root of a negative number is not a real number, the function does not have a value at (x=-12).

Exercise (PageIndex{5})

For the function (f(x)=sqrt[4]{3 x+4}), find

  1. (f(4))
  2. (f(-1))
Answer
  1. (f(4)=2)
  2. (f(-1)=1)

Exercise (PageIndex{6})

For the function (g(x)=sqrt[4]{5 x+1}), find

  1. (g(16))
  2. (g(3))
Answer
  1. (g(16)=3)
  2. (g(3)=2)

Find the Domain of a Radical Function

To find the domain and range of radical functions, we use our properties of radicals. For a radical with an even index, we said the radicand had to be greater than or equal to zero as even roots of negative numbers are not real numbers. For an odd index, the radicand can be any real number. We restate the properties here for reference.

Properties of (sqrt[n]{a})

When (n) is an even number and:

  • (a geq 0), then (sqrt[n]{a}) is a real number.
  • (a<0), then (sqrt[n]{a}) is not a real number.

When (n) is an odd number, (sqrt[n]{a}) is a real number for all values of (a).

So, to find the domain of a radical function with even index, we set the radicand to be greater than or equal to zero. For an odd index radical, the radicand can be any real number.

Domain of a Radical Function

When the index of the radical is even, the radicand must be greater than or equal to zero.

When the index of the radical is odd, the radicand can be any real number.

Example (PageIndex{4})

Find the domain of the function, (f(x)=sqrt{3 x-4}). Write the domain in interval notation.

Solution:

Since the function, (f(x)=sqrt{3 x-4}) has a radical with an index of (2), which is even, we know the radicand must be greater than or equal to (0). We set the radicand to be greater than or equal to (0) and then solve to find the domain.

Solve.

The domain of (f(x)=sqrt{3 x-4}) is all values (x geq frac{4}{3}) and we write it in interval notation as (left[frac{4}{3}, infty ight)).

Exercise (PageIndex{7})

Find the domain of the function, (f(x)=sqrt{6 x-5}). Write the domain in interval notation.

Answer

(left[frac{5}{6}, infty ight))

Exercise (PageIndex{8})

Find the domain of the function, (f(x)=sqrt{4-5 x}). Write the domain in interval notation.

Answer

(left(-infty, frac{4}{5} ight])

Example (PageIndex{5})

Find the domain of the function, (g(x)=sqrt{frac{6}{x-1}}). Write the domain in interval notation.

Solution:

Solve the function, (g(x)=sqrt{frac{6}{x-1}}) has a radical with an index of (2), which is even, we know the radicand must be greater than or equal to (0).

The radicand cannot be zero since the numerator is not zero.

For (frac{6}{x-1}) to be greater than zero, the denominator must be positive since the numerator is positive. We know a positive divided by a positive is positive.

We set (x-1>0) and solve.

(x-1>0)

Solve.

(x>1)

Also, since the radicand is a fraction, we must realize that the denominator cannot be zero.

We solve (x-1=0) to find the value that must be eliminated from the domain.

(x-1=0)

Solve.

(x=1) so (x/neq 1) in the domain.

Putting this together we get the domain is (x>1) and we write it as ((1, infty)).

Exercise (PageIndex{9})

Find the domain of the function, (f(x)=sqrt{frac{4}{x+3}}). Write the domain in interval notation.

Answer

((-3, infty))

Exercise (PageIndex{10})

Find the domain of the function, (h(x)=sqrt{frac{9}{x-5}}). Write the domain in interval notation.

Answer

((5, infty))

The next example involves a cube root and so will require different thinking.

Example (PageIndex{6})

Find the domain of the function, (f(x)=sqrt[3]{2 x^{2}+3}). Write the domain in interval notation.

Solution:

Since the function, (f(x)=sqrt[3]{2 x^{2}+3}) has a radical with an index of (3), which is odd, we know the radicand can be any real number. This tells us the domain is any real number. In interval notation, we write ((-infty, infty)).

The domain of (f(x)=sqrt[3]{2 x^{2}+3}) is all real numbers and we write it in interval notation as ((-infty, infty)).

Exercise (PageIndex{11})

Find the domain of the function, (f(x)=sqrt[3]{3 x^{2}-1}). Write the domain in interval notation.

Answer

((-infty, infty))

Exercise (PageIndex{12})

Find the domain of the function, (g(x)=sqrt[3]{5 x-4}). Write the domain in interval notation.

Answer

((-infty, infty))

Graph Radical Functions

Before we graph any radical function, we first find the domain of the function. For the function, (f(x)=sqrt{x}), the index is even, and so the radicand must be greater than or equal to (0).

This tells us the domain is (x≥0) and we write this in interval notation as ([0,∞)).

Previously we used point plotting to graph the function, (f(x)=sqrt{x}). We chose (x)-values, substituted them in and then created a chart. Notice we chose points that are perfect squares in order to make taking the square root easier.

Once we see the graph, we can find the range of the function. The (y)-values of the function are greater than or equal to zero. The range then is ([0,∞)).

Example (PageIndex{7})

For the function (f(x)=sqrt{x+3}),

  1. find the domain
  2. graph the function
  3. use the graph to determine the range

Solution:

  1. Since the radical has index (2), we know the radicand must be greater than or equal to zero. If (x+3 geq 0), then (x geq-3). This tells us the domain is all values (x geq-3) and written in interval notation as ([-3, infty)).
  2. To graph the function, we choose points in the interval ([-3, infty)) that will also give us a radicand which will be easy to take the square root.

c. Looking at the graph, we see the (y)-values of the function are greater than or equal to zero. The range then is ([0, infty)).

Exercise (PageIndex{13})

For the function (f(x)=sqrt{x+2}),

  1. find the domain
  2. graph the function
  3. use the graph to determine the range
Answer
  1. domain: ([-2, infty))


  2. Figure 8.7.3
  3. range: ([0, infty))

Exercise (PageIndex{14})

For the function (f(x)=sqrt{x-2}),

  1. find the domain
  2. graph the function
  3. use the graph to determine the range
Answer
  1. domain: ([2, infty))


  2. Figure 8.7.4
  3. range: ([0, infty))

In our previous work graphing functions, we graphed (f(x)=x^{3}) but we did not graph the function (f(x)=sqrt[3]{x}). We will do this now in the next example.

Example (PageIndex{8})

For the function, (f(x)=sqrt[3]{x}),

  1. find the domain
  2. graph the function
  3. use the graph to determine the range

Solution:

a. Since the radical has index (3), we know the radicand can be any real number. This tells us the domain is all real numbers and written in interval notation as ((-infty, infty))

b. To graph the function, we choose points in the interval ((-infty, infty)) that will also give us a radicand which will be easy to take the cube root.

c. Looking at the graph, we see the (y)-values of the function are all real numbers. The range then is ((-infty, infty)).

Exercise (PageIndex{15})

For the function (f(x)=-sqrt[3]{x}),

  1. find the domain
  2. graph the function
  3. use the graph to determine the range
Answer
  1. domain: ((-infty, infty))


  2. Figure 8.7.6
  3. range: ((-infty, infty))

Exercise (PageIndex{16})

For the function (f(x)=sqrt[3]{x-2}),

  1. find the domain
  2. graph the function
  3. use the graph to determine the range
Answer
  1. domain: ((-infty, infty))


  2. Figure 8.7.7
  3. range: ((-infty, infty))

Access these online resources for additional instruction and practice with radical functions.

  • Domain of a Radical Function
  • Domain of a Radical Function 2
  • Finding Domain of a Radical Function

Key Concepts

  • Properties of (sqrt[n]{a})
    • When (n) is an even number and:
      (a≥0), then (sqrt[n]{a}) is a real number.
      (a<0), then (sqrt[n]{a}) is not a real number.
    • When (n) is an odd number, (sqrt[n]{a}) is a real number for all values of (a).
  • Domain of a Radical Function
    • When the index of the radical is even, the radicand must be greater than or equal to zero.
    • When the index of the radical is odd, the radicand can be any real number.

Glossary

radical function
A radical function is a function that is defined by a radical expression.

Physiological role of antioxidants in the immune system

Diets contain naturally occurring antioxidant compounds that can stabilize highly reactive, potentially harmful molecules called free radicals. Free radicals are generated during normal cellular metabolism and result from the metabolism of certain drugs or xenobiotics. Exposure to UV light, cigarette smoke, and other environmental pollutants also increases the body's free radical burden. The harmful activities of free radicals are associated with damage to membranes, enzymes, and DNA. The ability of antioxidants to destroy free radicals protects the structural integrity of cells and tissues. This review focuses on data indicating that the functions of the human immune system depend on the intake of micronutrients, which can act as antioxidants. Recent clinical trials have found that antioxidant supplementation can significantly improve certain immune responses. Specifically, supplementation with vitamins C, E, and A or beta-carotene increased the activation of cells involved in tumor immunity in the elderly. Supplementation with the antioxidant vitamins also protected immune responses in individuals exposed to certain environmental sources of free radicals. Supplementation with vitamin A, a relatively weak antioxidant, decreases morbidity and mortality associated with measles infections in children.


Oxidation

Antioxidants are compounds that protect cells from damage caused by oxidation. But what is oxidation? Oxidation is a chemical reaction in which atoms lose electrons. Atoms have a nucleus (central core) which is positively charged. Orbiting around the nucleus are electrons which are negatively charged. The opposite attraction between the positively charged nucleus and negatively charged electrons keeps an atom stable. However, during metabolic reactions, atoms exchange electrons. Oxidation is the loss of electrons from an atom. Conversely, reduction is the gain of electrons by an atom. Oxidation and reduction usually occur together as an exchange reaction. One way to remember the difference between oxidation and reduction in the exchange reaction is to remember "OIL RIG":

  • OIL = Oxidation Is Loss of electrons
  • RIG = Reduction Is Gaining of electrons

FREE RADICALS

A free radical can be defined as any molecular species capable of independent existence that contains an unpaired electron in an atomic orbital. The presence of an unpaired electron results in certain common properties that are shared by most radicals. Many radicals are unstable and highly reactive. They can either donate an electron to or accept an electron from other molecules, therefore behaving as oxidants or reductants.[5] The most important oxygen-containing free radicals in many disease states are hydroxyl radical, superoxide anion radical, hydrogen peroxide, oxygen singlet, hypochlorite, nitric oxide radical, and peroxynitrite radical. These are highly reactive species, capable in the nucleus, and in the membranes of cells of damaging biologically relevant molecules such as DNA, proteins, carbohydrates, and lipids.[6] Free radicals attack important macromolecules leading to cell damage and homeostatic disruption. Targets of free radicals include all kinds of molecules in the body. Among them, lipids, nucleic acids, and proteins are the major targets.

Production of free radicals in the human body

Free radicals and other ROS are derived either from normal essential metabolic processes in the human body or from external sources such as exposure to X-rays, ozone, cigarette smoking, air pollutants, and industrial chemicals.[3] Free radical formation occurs continuously in the cells as a consequence of both enzymatic and nonenzymatic reactions. Enzymatic reactions, which serve as source of free radicals, include those involved in the respiratory chain, in phagocytosis, in prostaglandin synthesis, and in the cytochrome P-450 system.[7] Free radicals can also be formed in nonenzymatic reactions of oxygen with organic compounds as well as those initiated by ionizing reactions.

Some internally generated sources of free radicals are[8]

Some externally generated sources of free radicals are:

Free radicals in biology

Free radical reactions are expected to produce progressive adverse changes that accumulate with age throughout the body [ Table 1 ]. Such “normal” changes with age are relatively common to all. However, superimposed on this common pattern are patterns influenced by genetics and environmental differences that modulate free radical damage. These are manifested as diseases at certain ages determined by genetic and environmental factors. Cancer and atherosclerosis, two major causes of death, are salient 𠇏ree radical” diseases. Cancer initiation and promotion is associated with chromosomal defects and oncogene activation. It is possible that endogenous free radical reactions, like those initiated by ionizing radiation, may result in tumor formation. The highly significant correlation between consumption of fats and oils and death rates from leukemia and malignant neoplasia of the breast, ovaries, and rectum among persons over 55 years may be a reflection of greater lipid peroxidation.[9] Studies on atherosclerosis reveal the probability that the disease may be due to free radical reactions involving diet-derived lipids in the arterial wall and serum to yield peroxides and other substances. These compounds induce endothelial cell injury and produce changes in the arterial walls.[10]

Table 1


10-7A Orientation of Addition

The direction of addition of hydrogen bromide to propene clearly depends on which end of the double bond the bromine atom attacks. The important question is which of the two possible carbon radicals that may be formed is the more stable, the 1-bromo-2-propyl radical, (5), or the 2-bromo-1-propyl radical, (6):

From (ce) bond-dissociation energies of alkanes (see Table 5-6), the ease of formation and stabilities of the carbon radicals is seen to follow the sequence tertiary (>) secondary (>) primary. By analogy, the secondary 1-bromo-2-propyl radical, (5), is expected to be more stable and more easily formed than the primary 2-bromo-1-propyl radical, (6). The product of radical addition should be, and indeed is, 1-bromopropane:

Other reagents, such as the halogens, also can add to alkenes and alkynes by both radical-chain and ionic mechanisms. Radical addition usually is initiated by light, whereas ionic addition is favored by low temperatures and no light. Nevertheless, it often is difficult to keep both mechanisms from operating at the same time. This is important even when the alkene is symmetrical because, although the adduct will then have the same structural formula regardless of mechanism, the stereochemical configurations may differ. Electrophilic addition of halogens generally is a stereospecific antarafacial addition, but radical-chain additions are less stereospecific.

There are many reagents that add to alkenes only by radical-chain mechanisms. A number of these are listed in Table 10-3. They have in common a relatively weak bond, (ce), that can be cleaved homolytically either by light or by chemical initiators such as peroxides. In the propagation steps, the radical that attacks the double bond does so to produce the more stable carbon radical. For addition to simple alkenes and alkynes, the more stable carbon radical is the one with the fewest hydrogens or the most alkyl groups at the radical center.

Table 10-3: Reagents that add to Alkenes by Radical-Chain Mechanisms

The principles of radical addition reactions of alkenes appear to apply equally to alkynes, although there are fewer documented examples of radical additions to triple bonds. Two molecules of hydrogen bromide can add to propyne first to give cis-1-bromopropene (by antarafacial addition) and then 1,2-dibromopropane:


How to Reduce Free Radicals in Your Body

Reducing free radicals in your body includes both reducing the chance they will form and providing your body with antioxidants. The body produces antioxidants itself, but not in sufficient quantities alone. It's important to note, however, that since free radicals are produced during normal cellular processes, people may "do everything right" and still develop cancer.

Reducing your exposure to free radicals include both avoiding their sources (carcinogens) and providing your body with healthy antioxidants in your diet.

Lifestyle measures to reduce exposure include not smoking, avoiding processed foods, practicing caution with any chemicals you work with at home or on the job, and more.

As far as obtaining a healthy variety of antioxidants in your diet, experts in nutrition often recommend eating a "rainbow of foods" with different color foods often containing different classes of antioxidants.


3.2 Instrument sensitivity

The possible dependence of the HO2 detection sensitivity on the concentration of the gaseous water vapour mixing ratio was studied using two different radical sources. The water-dependent calibration factor is defined by Eq. (1), where c represents the instrument sensitivity that depends on the water concentration.

One of the radical sources is described in Sect. 2.4. Keeping the UV flux of the photolysis lamp constant, different HO2 concentrations were produced by varying the water-vapour mixing ratio between 0.1 % and 1.2 %. As the HO2 concentration provided by the calibration can be accurately calculated for different water mixing ratios, the influence of water on the HO2 detection sensitivity could be investigated. Measurements under dry conditions were not possible, because the calibration source needs water to generate HO2 .

For low water-vapour concentrations, ozonolysis of 2,3-dimethyl-2-butene was used as a radical source. For that purpose, the alkene was added in a concentration of 30 ppbv to a mix of synthetic air and 200 ppbv ozone. The radical source (with photolysis lamp switched off) was used as a flow tube to overflow the inlet of the instrument with this gas mixture. A total of 0.2 % CO was added to scavenge OH radicals produced from the ozonolysis reaction by a fast conversion of OH to HO2 . The water mixing ratio was altered during the ozonolysis experiment from 0.0 % to 0.6 %. Assuming that the HO2 concentration from the ozonolysis is constant, the relative change in the signal gives the relative change of the instrument sensitivity. Absolute sensitivities were derived by scaling the HO2 signals from the ozonolysis experiment to the concentration derived by the water-dependent calibration from the radical source by multiplication with a constant factor.

Figure 3Measured HO2 sensitivity as a function of the water mixing ratio in two experiments. For the calibration, HO2 was produced by the radical source while varying the water-vapour concentration which causes a change in the HO2 radical concentration. During the ozonolysis experiment, HO2 was produced from the ozonolysis of 2,3-dimethyl-2-butene, which is independent of the water vapour mixing ratio. The red line shows a third-order polynomial fit applied to the calibration data for the range of water vapour mixing ratios higher than 0.1 %.

Figure 3 shows the sensitivity determined for each water vapour mixing ratio, showing a decreasing sensitivity with increasing water vapour mixing ratio for atmospheric relevant water mixing ratios higher than 0.1 % . The water-dependent decrease in sensitivity is nearly linear for this range of water vapour mixing ratios. For water vapour mixing ratios of less than 0.1 % , the sensitivity drops quickly by a factor of 7 under dry conditions compared to the maximum sensitivity at a 0.1 % water vapour mixing ratio.

Two effects contribute to the water dependence: the initial increase of sensitivity (below 0.1 % H2O ) comes from the stabilising effect of H2O . Br − adds H2O , forming a loosely bound complex of H 2 O ⋅ Br - then, the H 2 O ⋅ Br - complex reacts with HO2 according to the forward Reaction (R2). The steady decrease of sensitivity by a factor of 2 when the H2O mixing ratio is further increased to 1.2 % comes from the back reaction of Reaction (R2).

The water vapour dependence of the sensitivity can be parameterised by a third-order polynomial (Eq. 2) for water vapour mixing ratios higher than 0.1 % . This is typically sufficient for atmospheric conditions. At lower water vapour mixing ratios the parameterisation in Eq. (3) provides a good approximation. Such low water vapour mixing ratios were present in the chamber experiments after flushing the chamber before an experiment started.

S is the signal normalised by the primary ion, a , b , c , d are the fit parameters, and H2O is the absolute water vapour mixing ratio. During the series of chamber experiments presented in Sect. 3.5, calibrations were done in between the experiments. In the middle of the series of experiments (6 June), settings of the instrument were tuned, changing the sensitivity of the instrument. In total six calibrations were performed.

To gain sensitivity, the wall contact was reduced by directly sampling via a nozzle into the ion flow tube in the instrument used here. The ion flow tube was further optimised for length and pressure to improve the sensitivity for HO2 . Basically, the ion flow tube used during this study (130 mm length) was compared to a similar ion flow tube with a length of 200 mm. However, this resulted in 50 % less sensitivity at 120 hPa, which has been identified as the optimal pressure in terms of sensitivity. Finally, the flows were optimised to gain the maximum amount of sensitivity. Further sensitivity can be gained by combining both isotopic signals for the data analysis, as already mentioned by Sanchez et al. (2016) .

For the chamber experiments, the chamber air was humidified at the beginning of each experiment. At that time, no HO2 is expected to be present in the chamber. Therefore, the signal caused by the constant HO2 background changes with the water vapour dependence of the instrument sensitivity (see next section) and could be used to determine the relative change of the sensitivity of water vapour for an individual experiment during this measurement campaign. All HO2 data from the chamber experiments shown in Sect. 3.5 were evaluated by applying this procedure.

As shown in Fig. 3, the instrument response to the change of the water-vapour concentration is similar to both methods of radical production. In addition, the instrument's sensitivity under dry conditions could be tested in the ozonolysis case showing that the instrument sensitivity drops by nearly an order of magnitude in the absence of water vapour. Because of the fast drop of the instrument's sensitivity for water vapour mixing ratios below 0.1 % , it is beneficial to add water vapour to the ion flow tube under very dry conditions of sampled air to maintain a high instrument sensitivity.

Sanchez et al. (2016) used a similar approach to calibrate their instrument via photolysis of water, but they used water mixing ratios in the pptv range to keep HO2 concentrations in an atmospheric range. They used purified air for the calibration source. This study uses synthetic air (purity 99.9999 %).

Sanchez et al. (2016) found a constant sensitivity for water vapour mixing ratios between 0.2 % and 0.8 %, whereas a 30 % decrease is observed here. Only for one sensitivity measurement at a 0.06 % water mixing ratio is an increased sensitivity of approximately 50 % reported by Sanchez et al. (2016) . The reason for this different behaviour is not clear, but one may speculate that the design of the ion flow tube and inlet nozzle might impact the collision probability of ion clusters. The relative change of the instrument's sensitivity as conditions become drier is not reported in Sanchez et al. (2016) , so it is not clear if the sensitivity drops for dry conditions in their instruments as observed here.


Kanji Radicals Can Indicate Meaning

Similar to English, where we have prefixes, suffixes, infixes, and roots, all of which indicate a modicum of meaning, a kanji’s radical can sometimes give hints as to the kanji’s meaning. Now, this is not nearly as reliable of a general rule as it is for English words, as you will see, but it can still be useful for giving your memory a sort of “hook” by which you can remember each kanji.

Let’s look at another common radical – ( みず ) , which means “water”. Now, this radical is pretty interesting, because it actually appears in different forms: in addition to the standalone form 水, it can also appear in the form of three strokes on the left side 氵. There are a few other radicals that do this. In fact, we already saw that the 人 radical (“person”) also appears in the form 亻.

Now we know that 氵 means “water,” let’s look at a few kanji that use it to see if we can spot a few similarities in meaning:

  • ( かわ) – “river”
  • ( なみだ) – “tears”
  • ( さけ) – “alcohol”
  • ( いけ) – “pond”

… and the list goes on. Unfortunately, though numerous, these examples are cherry-picked. There are plenty of other kanji that use the water radical that no longer carry much (if any) connotation to water anymore:

  • ( けつ) – “decision”
  • ( ) – “government cure”
  • ( はく) – “overnight”
  • ( ほう) – “law”

So, although the kanji radicals originally had a lot more significant meaning to them, unfortunately, you can no longer rely on them to be related to the practical meaning of the kanji.

Kanji Radicals Can Indicate Reading

Similarly to how some components can hint at the meaning of a kanji, some other components indicate reading instead of meaning. Again, this is not a hard and fast rule, but for some components it works surprisingly well.

An important note: in this category, many of the reading indicators are compound components, i.e. they are made of multiple sub-components.

Let’s start with 复, which is not really used by itself, but when it appears within another kanji, that one tends to be pronounced フク: ( ふく ) , ( ふく ) , ( ふく ) . If a kanji contains 生, it is more often than not pronounced せい: ( せい ) , ( せい ) , ( せい ) . There’s a myriad of these so-called “phonetic components,” an excellent write-up of which can be found here .

Wanna know why the pronunciations of the 复 and 生 are represented in different ways? Read this article to learn the difference between hiragana and katakana.


Supporting Information Available

Tables of LFP kinetic results for radicals 2, 3, and 5, synthetic details for the preparation of radical precursors, synthetic details for preparation of the requisite amides, and NMR spectra of new compounds (58 pages, print/PDF). This material is contained in libraries on microfiche, immediately follows this article in the microfilm version of the journal, and can be ordered from the ACS. See any current masthead page for ordering information and Web access instructions.


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