## Info

Figure 5.14 A function with different left-hand and right-hand derivatives I

Example 5.2 Typically the marginal rate of income tax varies across income ranges. For example.

suppose that the first \$5,000 income earned is not taxed at all. the next \$10,000 earned is taxed at the rate of 15%, and any further income is taxed at the rate of 25%. Let T(y) represent the income tax schedule for this example. The simplest way to graph this function is to think about how the marginal tax rate changes over the various income ranges:

Marginal tax rate =

0, for 0 < y S 5.000 0.15. for 5,000 < y < 15,000 0.25. for v > 15,000

Since no lax is paid on zero income, we have T(0) = 0. Knowing the rate at which the tax increases allows us to draw the tax function (see figure 5.15). Note that the tax function is not discontinuous. However, it is not differentiable at the points y = 5,000 or v - 15.000 because the marginal tax rate changes at these points, with the result that the left- and right-hand derivatives are not the same. Notice that it we graph the derivative function. 7"(y), on the intervals [0. 5.000], (5,000, 15,0001. and (15,000. oo), as is done in figure 5.16, we see that it is discontinuous at the points where 7"(y) is not differentiable This result illustrates that since the left- and right-hand limits of the derivative are not equal at a point of nondifferentiability. the derivative function is not defined at that point.

marginal tax rate = (J.25

marginal tax rate = 0.15

5,000 10,000 15.000 25,000 .v Figure 5.15 Income tax schedule for example 5.2

marginal tax rate = (J.25

marginal tax rate = 0.15

5,000 10,000 15.000 25,000 .v Figure 5.15 Income tax schedule for example 5.2

5,000 10,000 15,000 25,000 v

Figure 5.16 Marginal income tax function for example 5.2

The tax function. T(y), is defined in equation (5.4). The first part of the function. T(y) = 0, forO < y < 5,000 reflects the fact that the first \$5,000 earned is not taxed and so anyone earning less than this amount pays zero tax The second part. T(y) = 0.15(y - 5,000), for y t (5,000, 15.000] indicates that any income earned in excess of \$5,000, but less than \$15,000. is taxed at the rate of 15%, The third part, T{y) = 0.15(y - 5,000) + 0.10(y - 15,000). for y > \$15,000 indicates that income earned in excess of \$15,000 is taxed at a rate of 25% and so an additional 0.l0(y - 15,000) must be added to the 15% which is charged on income earned over \$5,000.

This tax schedule can be simplified to get

0 < y < 5.000 5.000 < y < 15,000 v > 15.000

0.15.v - 750. 5.000 < y < 15.000 0.25y- 2,250. y > 15,000

Returning to the definition for differentiability (definition 5.5). we can see a close relationship to the condition for continuity (refer to chapter 4.1). For the function / to be differentiate at the point x, the lim^,_u f(x + Ax.) must exist; that is, the left-hand limit, limA.,-.o f(x + Ax), must equal the right-hand limit, limAv_o* f(x + Ax). Moreover, as we noted earlier in this section, it must also be the case that lim^J_o fix + Ax) = /(x). These conditions imply that the function f(x) is continuous. Thus, if a function is not continuous, then it is not differentiable and so we have demonstrated the following theorem.

Theorem 5.1 If f\x) exists (i.e.. the function fix) is differentiable) at the point r = n. then the function /(a) must also be continuous at this point.

It is important to note that continuity is a necessary but not a sufficient con- ■ dition for differentiability. In other words, a function may be continuous at sort«' point yet not differentiable there. This is clearly established by example 5.2 above. At the point x = l, we find that lim f(x + Ax) = I, lim fix + Ax) = 1, /'(I) = I

which implies that the function is continuous. However, it is not differentiable a (he point x = 1.

We can employ the concepts of left- and right-hand derivatives in defining whether a function is differentiable in the case in which the domain of the function is a closed interval, |a. or has only one boundary point. [u, +oc) or (—00, b]. Note that if the domain of a function is the closed interval [a, ft], then the question arises of defining its differentiability at a and b. To handle such cases, we ask whether the relevant one-sided limits exist as one approaches x = a from the right or v = b from the left. The following definition for differentiability of a function defined on a closed interval [a, b\can be extended in an obvious way for functions defined on the sets [«. 00) and (—00, b\.

A function /'(.r) defined on the domain x € fc/. b] is differentiable on b\ if (i) the right-hand derivative for f(x) exists at x = a. (ii) the left-hand derivative exists at a = b. and (iii) fix) is differentiable at every point in the open set (a, b).

Example 5.3

Figure 5.17 Function with a domain being a closed interval

An example of a situation in which it is natural to Use a function defined on a closed interval is the case of trade quotas. If a firm can export up to a maximum amount, say b, of some product into a country, then one would define sales revenue for this activity by ftu ), x € (0. b\ Sales.jc, must not exceed the value b and cannot be negative.

### Modeling Enforcement as a Probability

Consider an enforcement agency choosing a level of enforcement of a law against some crime (e.g., tax evasion). We can represent the level of enforcement by the probability with which the crime will be detected. Let p be this probability of detection By definition, p must lie in the closed interval ((I. 11. If ihe function c(p) represents the resource cost of achieving detection level p. then p e [0. 1] is the domain of the function. For example, c(p) = kp2 (see figure 5.17) could represent an enforcement cost function which has right-hand derivative of value zero at p = 0 and left-hand derivative of value 2k at p = 1. ■

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