3. Análisis de la Ejecución Presupuestaria de la Administración Nacional
3.1. Análisis de la ejecución de los ingresos, gastos y resultados
we can see how the relative change in quantity is simply related to the relative change in price via the elasticity of demand.
Example 3.20 Suppose that the demand function for some good is given by qD(p) = 10p−r where r is a constant. Find the elasticity of demand. What does this tell us about the effect of relative changes in price on relative changes in quantity?
Here we have q = qD(p) = 10p−r where r is a constant and so, q0(p) =−10rp−r−1,
which means that the elasticity of demand is given by ε(p) =−p
qq0(p) =− p 10p−r
− 10rp−r−1
= r, i.e. ε(p) is a constant too. This means that, using
∆q
q ' −ε(p)∆p p ,
we see that a relative increase in price of, say, x% will lead to a relative decrease in quantity purchased of (approximately) rx%.
Indeed, we will see, in Section 4.2.3, that elasticities can also give us useful information about how the revenue, R = pq, generated from selling a quantity, q, at a price of p per unit will be affected by increases in the price.
3.3.4 Existence of derivatives
Although we will usually be dealing with situations where a function has a derivative at every point where it is defined, we will occasionally encounter situations where there is at least one point at which the derivative of a function does not exist. Just so that we are aware of what this means and the kinds of situation in which it can arise, we consider some of the most common ways in which a derivative can fail to exist at a certain point.8
Discontinuous functions
If a function is discontinuous at a point, i.e. there is a point at which the function is not continuous, then the derivative will not exist at that point as the next example
illustrates.
Example 3.21 Show that the derivative of the function defined by
f (x) =
(1 x≥ 0
−1 x < 0,
8See, for example, Section 2.8 of Binmore and Davies (2002) for a discussion of some similar cases.
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3.3. Using derivatives
does not exist when x = 0.
This function is illustrated in Figure 3.5(a) and, clearly, as the function is a
continuous horizontal line when x 6= 0, its derivative is defined and equal to zero as long as x6= 0. However, when x = 0, the function is discontinuous and its derivative, if it exists, would be given by
f0(0) = lim
h→0
f (h)− f(0)
h .
However, here we can not just find
f (h)− f(0)
h ,
and let h→ 0 as we did in Section 3.1 since the value of f(h) is different depending on whether h is positive or negative. In such cases, we say that the limit we seek, i.e.
hlim→0
f (h)− f(0)
h ,
exists if, firstly, both of the limits
hlim→0−
exist9 and, secondly, if they exist, they must be equal. But, using the given function, we see that
i.e. neither of these limits exists as ‘∞’ is not a value10 but more of a notational convenience which tells us that a function is getting arbitrarily large in the limit.
Consequently, we see that
f0(0) = lim
h→0
f (h)− f(0)
h ,
fails to exist too and so the derivative of this function does not exist at x = 0.
Of course, the graph of a function can also have a discontinuity due to the presence of a vertical asymptote. In such cases, the function is not actually defined at the value of x where the asymptote occurs and so, because of this, the derivative cannot exist at this point either.11 In both of these cases, as we can’t ascribe a gradient to the function at these points, the function can’t have a tangent line at these points.
9Notice that the former limit allows us to deal with negative h and the latter allows us to deal with positive h. Also recall that the notation h→ 0− and h→ 0+ was explained in Example 2.2.
10That is, it is not a real number.
11We’ll come across this again in Section 4.4.3.
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Figure 3.5:The graphs of three functions that have no derivative at x = 0 as explained in (a) Example 3.21, (b) Example 3.22 and (c) Example 3.23. We note however that, unlike the functions in (a) and (b), the function in (c) does have a tangent line at x = 0 given by the vertical line with equation x = 0.
Continuous functions with ‘corners’
But, even if a function is continuous at every point, the derivative will not exist at points where the curve changes too sharply, i.e. when the curve has a ‘corner’, as the next example illustrates.
Example 3.22 Show that the derivative of the function f (x) =|x| does not exist when x = 0.
This function is illustrated in Figure 3.5(b) and, clearly, as the function is the continuous straight line f (x) =−x when x < 0 and f(x) = x when x > 0, its
derivative is defined and equal to −1 when x < 0 and 1 when x > 0. However, when x = 0, the function has a ‘corner’ and its derivative, if it exists, would be given by
f0(0) = lim
h→0
f (h)− f(0)
h .
However here, as in Example 3.21, we can not just find f (h)− f(0)
h ,
and let h→ 0 as we did in Section 3.1 since the value of f(h) is different depending on whether h is positive or negative. In such cases, we again say that the limit we seek, i.e.
hlim→0
f (h)− f(0)
h ,
exists if, firstly, both of the limits
h→0lim−
exist12 and, secondly, if they exist, they must be equal. But, using the given function, we see that
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3.3. Using derivatives
and
h→0lim+
f (h)− f(0)
h = lim
h→0+
h− 0
h = lim
h→0+1 = 1,
i.e. both of these limits exist, but they are clearly not equal. Consequently, we see that
f0(0) = lim
h→0
f (h)− f(0)
h ,
fails to exist and so the derivative of this function does not exist at x = 0.
Observe that, in this case, the limits as h→ 0+ and as h→ 0− both exist, but the problem occurs because they are not equal and so we cannot ascribe a value to the derivative (i.e. the limit as h→ 0) in such situations. In particular, as this means that we can’t ascribe a gradient to f at this point, the function can’t have a tangent line here either.
Continuous functions with ‘vertical tangent lines’
Also, if a function is continuous at every point, the derivative will not exist at points where the gradient of the curve becomes ‘infinite’, i.e. when the curve has a ‘vertical tangent line’, as the next example illustrates.
Example 3.23 Show that the derivative of the function f (x) = x1/3 does not exist when x = 0.
This function is illustrated in Figure 3.5(c) and, clearly, we can see that its derivative is given by
f0(x) = 13x−2/3= 1 3x2/3,
which exists as long as x6= 0. Of course, when x = 0, the derivative cannot exist since, if we were to use this formula, we would have to ‘divide by zero’ and this is never allowed. However, we can see from Figure 3.5(c) that the graph of the function has a vertical tangent line at x = 0 which is given by the vertical line with equation x = 0.13 Thus, we have a situation where the derivative of the function does not exist at x = 0, but it does have a tangent line at that point.
Observe that, in cases where the tangent line to f at a point is a vertical line we cannot use (3.1) to find its equation as its derivative is not defined.14
12Again, as in Example 3.21, the former limit allows us to deal with negative h and the latter allows us to deal with positive h.
13Notice that the tangent lines of the function are getting steeper as we move towards x = 0 on the left and shallower as we move away from x = 0 on the right.
14We’ll come across this again in Section 4.4.3.