1 Linear Approximation and Applications

Math 20A - Calculus by Jon Rogawski Chapter 4 - Applications of the derivative Prepared by Jason Gaddis

1 Linear approximation and applications

Remark 1.1. We can use tangent lines to approximate a curve at a specific point. If we allow for a small amount of error, we can approximate the curve on some interval The error grows as we make the interval larger. This is called linear approximation. Definition 1.2. If f is differentiable at x = a and ∆x is small, then

∆f ≈ f 0(a)∆x where ∆x = f(a+δx)−f(a) is the linear approximation of ∆f (or tangent line approximation). √ Example 1.3. Use Linear approximation to estimate 100.2 − 10.

√ 0 1 −1/2 Let f(x) = x at x = 100 and ∆x = .2. Then f (x) = 2x , so ∆f is approximated by 1 1 f 0(100)∆x = (.02) = .001. 2 10 √ A calculator gives 100.2 = 10.009995005, so the error in the diﬀerence is

|009995005 − .001| = 0.008995005.

Remark 1.4 (Differential notation). If y = f(x), where f is a differentiable function, then the differential dx is an independent variable. The differential dy is then defined in terms of dx by the equation

dy = f 0(a)dx.

Now, ∆y ≈ dy. Deﬁnition 1.5. If f is diﬀerentiable and x is close to a then the linearization of f at a is the linear function whose graph is the tangent line, that is, L(x) = f(a) + f 0(a)(x − a).

1 π Example 1.6. Find the linearization of the function f(x) = sin x at a = 6 π and use it to approximate sin( 5 ). √ 1 0 3 We have sin(π/6) = 2 and f (π/6) = cos(π/6) = 2 . Then the linearization of f at a is √ 1 3 π L(x) = + (x − ). 2 2 6 Then L(π/5) = 0.591. The actual value 0.588. Remark 1.7. Error is relatively meaningless by iteslf. To standardize error, we deﬁne the percentage error to be error × 100%. actual error Example 1.8. Compute the percentage error in the previous example.

The actual value was 0.588, so the error is |.591 − .588| = .003. Hence, the percentage error is

.003 × 100 = .510. .588 1 Thus, the error is approximately 2 of a percent.

2 2 Extreme Values

Definition 2.1. Let f(x) be a function on an interval I and let a ∈ I. We say that f(a) is the absolute maximum of f on I if f(x) ≤ f(a) for all x ∈ I. We say that f(a) is the absolute minimum of f on I if f(x) ≥ f(a) for all x ∈ I. Remark 2.2. The word “absolute” is sometimes replaced by “global” Remark 2.3. A function may not necessarily have an absolute maximum or minimum on some interval. Consider, for example, the function f(x) = x2 on any open interval. However, there is one case where we can guarantee the existence of absolute extrema. Theorem 2.4 (The Extreme Value Theorem). If f is continuous on a closed interval [a, b], then f attains an absolute maximum and absolute minimum on [a, b]. Remark 2.5. If either of the hypotheses above are removed, the theorem is no longer holds. Definition 2.6 (Local Extrema). A function f has a local maximum at x = c if f(c) ≥ f(x) for all x in some open interval containing c. Similarly, f has a local minimum at c if f(c) ≤ f(x) for all x in some open interval containing c. Remark 2.7. We can see relatively easily from a graph that local extrema occur at points where the graph of the function has a horizontal tangent line or where the derivative does not exist (vertical tangent line). Definition 2.8. A number c in the domain of f is a critical number if f 0(c) = 0 or f 0(c) does not exist. Example 2.9. Find the critical numbers of f(x) = x3 − x2 − x.

We have f 0(x) = 3x2 − 2x − 1 = (3x + 1)(x − 1). Thus, f 0(x) = 0 when x = −1/3 or x = 1. These are the critical points of f. Theorem 2.10 (Fermat’s Theorem). If f has a local maximum or minimum at c, then c is a critical point.

3 Remark 2.11. This is only one of many theorems attributed to Pierre de Fermat (1601-1665), who was a lawyer and amateur mathematician. Proof. Suppose f(c) is a local minimum. If f 0(c) does not exist, we are done, so assume it does exist. We claim f 0(c) = 0. Because f(c) is a local minimum, f(c + h) ≥ f(c) for small values of h 6= 0. That is, f(c + h) − f(c) ≥ 0. Then f(c + h) − f(c) f 0(c) = lim ≥ 0 h→0+ h f(c + h) − f(c) f 0(c) = lim ≤ 0. h→0− h Hence, f 0(c) ≥ 0 and f 0(c) ≤ 0, but this only holds if f 0(c) = 0. Remark 2.12. Consider f(x) = x3. This has a critical point at x = 0 but it is not a local max or min. Thus, while all local max and min occur at critical points, not all critical points yield local extrema. Example 2.13. In the previous example, determine whether the critical points are local max, local min, or neither.

We have f(x) = x3 − x2 − x, f 0(x) = 3x2 − 2x − 1 = (3x + 1)(x − 1), and critical points c = −1/3, x = 1. For x < −1/3, f 0(x) > 0, so the function is increasing. For −1/3 < x < 1, f 0(x) < 0, so the function is decreasing. Since the function switched from increasing to decreasing at x = −1/3, it is a local max. Similarly, for x > 1, f 0(x) > 0, so the function is increasing. Since the function switched from decreasing to increasing at x = 1, it is a local min. Theorem 2.14 (Extreme values on a closed interval). Assume that f(x) is continuous on [a, b] and let f(c) be the minimum or maximum value on [a, b]. Then c is either a critical point or one of the endpoints a or b. Example 2.15. Find the abs. max and abs. min values of f(x) = x3 −6x2 + 9x + 2 on the closed interval [−1, 4]. Theorem 2.16 (Rolle’s Theorem). Assume that f(x) is continuous on [a, b] and diﬀerentiable on (a, b). If f(a) = f(b), then there exists a number c between a and b such that f 0(c) = 0.

4 Example 2.17. Verify that the function f(x) = x3 −3x2 +2x+5 satisﬁes the hypotheses of Rolle’s Theorem on the interval [0, 2]. Then ﬁnd all numbers c that satisfy the conclusion of the theorem.

5 3 The Mean Value Theorem and Monotonicity

Theorem 3.1 (Rolle’s Theorem). Assume that f(x) is continuous on [a, b] and differentiable on (a, b). If f(a) = f(b), then there exists a number c between a and b such that f 0(c) = 0. Example 3.2. Verify that the function f(x) = x3 − 3x2 + 2x + 5 satisfies the hypotheses of Rolle’s Theorem on the interval [0, 2]. Then find all numbers c that satisfy the conclusion of the theorem. Example 3.3. Show that the equation 2x − 1 − sin x = 0 has exactly one real root. Theorem 3.4 (The Mean Value Theorem). Let f be a function that satisfies the following hypotheses: 1. f is continuous on the closed interval [a, b] 2. f is differentiable on the open interval (a, b) Then there is a number c in (a, b) such that f(b) − f(a) f 0(c) = . b − a Equivalently, f(b) − f(a) = f 0(c)(b − a). Example 3.5. Verify that the function f(x) = x3 + x − 1 satisfies the hypotheses of the Mean Value Theorem on the interval [0, 2]. Then find all numbers c that satisfy the conclusion of the theorem. Theorem 3.6. If f 0(x) = 0 for all x in the interval (a, b), then f is constant on (a, b). Corollary 3.7. If f 0(x) = g0(x) for all x in an interval (a, b), hen f − g is constant on (a, b); that is, f(x) = g(x) + c where c is a constant. Remark 3.8. We will see how the derivative gives us information about the behavior of a function. Our first result is a consequence of the MVT. Theorem 3.9 (The sign of the derivative). Let f be differentiable on an open interval (a, b).

6 (a) If f 0(x) > 0 on an interval, then f is increasing on that interval. (b) If f 0(x) < 0 on an interval, then f is decreasing on that interval.

0 Proof. Suppose f (x) > 0 for all x ∈ (a, b). Choose points x1 < x2 in (a, b). We claim f(x2) > f(x1). By the MVT, there exists a c ∈ (x1, x2) such that

0 f(x2) − f(x1) = f (c)(x2 − x1) > 0.

0 Therefore, f(x2) > f(x1). The case f (x) < 0 is similar. Example 3.10. Given f(x) = x4 − 4x − 1, ﬁnd the intervals on which f is increasing or decreasing. Theorem 3.11 (First Derivative Test). Suppose that c is a critical number of a continuous function f. • If f 0 changes from positive to negative at c, then f has a local maximum at c. • If f 0 changes from negative to positive at c, then f has a local minimum at c. • If f 0 does not change sign at c, then f has no local maximum or minimum at c. Example 3.12. Find the local minimum and maximum values of the function f in the previous example.

7 4 The shape of a graph

Remark 4.1. Now that we have seen what the first derivative can tell us, we turn our attention to the second derivative. Definition 4.2. If the graph of f lies above all of its tangents on an interval I, then it is is called concave upward (up) on I. If the graph of f lies below all of its tangents on I, it is called concave downard. A point P on a curve y = f(x) is called an inflection point if f is continuous there and the curve switches concavity there. Theorem 4.3 (Concavity Test). Assume that f 00(x) exists for all x ∈ (a, b), • If f 00(x) > 0 for all x ∈ I, then the graph of f is concave upward on I. • If f 00(x) < 0 for all x ∈ I, then the graph of f is concave downward on I. • If f 00(c) = 0 and f 00(x) changes sign at x = c, then f(x) has a point of inflection at x = c. Example 4.4. Given f(x) = x4 − 4x − 1, find the intervals on which f is concave up and concave down. Find any inflection points. Theorem 4.5 (Second Derivative Test). Let c be a critical point of f. If f 00(c) exists, then • If f 00(c) > 0, then f has a local minimum at c. • If f 00(c) < 0, the f has a local maximum at c. • If f 00(c) = 0, then the test is inconclusive.

8 Example 4.6. Given h(x) = 10x3 − x5, ﬁnd the following: (a) Intervals of increase or decrease. (b) Local max and min values. (c) Intervals of concavity and inﬂection points. Sketch the graph of the function using the information above.

9 5 L’Hˆospital’s Rule

Remark 5.1. We now move into studying asymptotic behavior of functions. A key tool in this study is a rule which allows us to compute limits of certain rational functions of indeterminate form. Theorem 5.2 (L’Hôspital’sRule). Assume that f(x) are differentiable on an open interval containing a and that f(a) = g(a) = 0. Also assume that g0(x) 6= 0 (except possibly at a). Then f(x) f 0(x) lim = lim x→a g(x) x→a g0(x) if the limit on the right exists or is infinite. This conclusion also holds if f(x) and g(x) are differentiable for x near (but not equal to) a and

lim f(x) = ±∞ and lim g(x) = ±∞. x→a x→a Furthermore, this rule is valid for one-sided limits. Proof. Assume f 0 and g0 are continuous at x = a and that g0(a) 6= 0. Then g(x) 6= g(a) for x near but not equal to a, and

f(x)−f(a) f(x) f(x) − f(a) = = x−a . g(x) g(x) − g(a) g(x)−g(a) x−a Now using the quotient law for limits,

f(x)−f(a) 0 0 f(x) limx→a f (a) f (x) lim = x−a = = lim . x→a g(x) g(x)−g(a) g0(a) x→a g0(x) limx→a x−a

sin x Example 5.3. Evaluate lim . x→0 x

Remark 5.4 (Indeterminate Diﬀerence). Suppose limx→a f(x) = ±∞ and limx→a g(x) = ±∞. We can evaluate lim(f − g)(x) by ﬁnding a common x→a denominator and writing as a single quotient. 1 1 Example 5.5. Evaluate lim − . x→1 ln x x − 1

10 Remark 5.6 (Assumptions matter). LR does not apply to all limits. Con- x − 1 sider lim . If we tried to apply LR, this would become lim 1 = 1, but x→2 x + 3 x→2 x − 1 1 clearly lim = . x→2 x + 3 5 Remark 5.7 (Indeterminate Powers). We consider lim[f(x)]g(x). There are x→a three forms, 1. lim f(x) = 0 and lim g(x) = 0 (type 00) x→a x→a 2. lim f(x) = ∞ and lim g(x) = 0 (type ∞0) x→a x→a 3. lim f(x) = 1 and lim g(x) = ±∞ (type 1∞) x→a x→a We evaluate by taking the natural log and evaluating the indeterminate product. Example 5.8. Evaluate lim xx x→0+ Theorem 5.9 (L’Hôspital’sRule for limits at infinity). Assume that f(x) and g(x) are differentiable in an interval (b, ∞) and that g0(x) 6= 0 for x > b. If lim f(x) and lim g(x) exist and either both are zero or both are infinite x→∞ x→∞ then f(x) f 0(x) lim = lim x→∞ g(x) x→∞ g0(x) if the limit on the right exists or is infinite. A similar result holds for limits as x → −∞. ex Example 5.10. Evaluate lim . x→∞ x3 ex Remark 5.11. In fact, a similar argument shows that lim = ∞ for all x→∞ xn n. This means that the function ex grows faster than that of xn.

11 6 Graph sketching and asymptotes

Remark 6.1. We have four basic shapes for graphs corresponding to increasing/decreasing and concavity. We call a point where the graph changes shape a transition point. We have learned how to graph based on this information. We want to add our knowledge on asymptotic behavior to the mix. Example 6.2. Sketch the graph of f(x) = xex.

x+3 Example 6.3. Sketch the graph of y = x−2.

12 7 Applied Optimization

Remark 7.1. We will utilize our techniques for graphing, specifically those skills needed to find absolute minima and maxima, to solve optimization problems. This has great application to areas of business and engineering, when we try to make systems as efficient as possible. In these problems, we will want to draw a diagram, when possible, and clearly define our variables. The key is to find the objective function which we wish to maximize and minimize. Example 7.2. Find two numbers whose difference is 100 and whose product is a minimum. (Hint: use the function f(x) = x(x − 100).) Example 7.3. Find the dimensions of a rectangle with area 1000m2 whose perimeter is as small as possible. Remark 7.4. Sometimes we do not have the luxury of working in a closed interval. Our domain may be (a, ∞) for some number a, or (−∞, b) for some number b. For this situation, we have the following modification of the First Derivative Test. Proposition 7.5 (First Derivative Test for Absolute Extreme Values). Sup- pose that c is a critical number of a continuous function f defined on an interval. (a) If f 0(x) > 0 for all x < c and f 0(x) < 0 for all x > c, then f(c) is the absolute maximum value of f. (b) If f 0(x) < 0 for all x < c and f 0(x) > 0 for all x > c, then f(c) is the absolute minimum value of f. Example 7.6. A box with a square base and open top must have a volume of 32000cm3. Find the dimensions of the box that minimize the amount of material used. Remark 7.7. Is it possible to maximize the amount of material used?

8 Newton’s Method

13 9 Antiderivatives

Remark 9.1. A sort of inverse operation to differentiation is antidifferenti- ation. The reason this is not a true inverse operation will become clear. Definition 9.2. A function F (x) is an antiderivative of f(x) on (a, b) if F 0(x) = f(x) on (a, b). Example 9.3. • F (x) = cos x is an antiderivative of f(x) = sin x.

1 2 • F (x) = 2x is an antiderivative of f(x) = x. 1 2 • F (x) = 2x + 1 is also an antiderivative of f(x) = x. Theorem 9.4 (The general antiderivative). Let F (x) and G(x) be antiderivatives of f(x) on (a, b), then G(x) = F (x) + C for some constant C. Proof. Let H(x) = G(x)−F (x), then H0(x) = F 0(x)−G0(x) = f(x)−f(x) = 0 and so H is constant on (a, b). Thus, H(x) = C, so G(x) = F (x) + C. Remark 9.5. In a sense, once we know one antiderivative, we know all of them. We denote by R f(x) dx the general antiderivative (or indeﬁnite integral) of f. That is, Z f(x) dx = F (x) + C, means F 0(x) = f(x). Theorem 9.6 (Power rule for integrals). For n 6= −1, Z xn+1 xn dx = + C. n + 1

1 Theorem 9.7. The function F (x) = ln |x| is an antiderivative of y = x in the domain {x | x 6= 0}. Theorem 9.8 (Linearity of the indeﬁnite integral). Let c be a constant. Z Z Z f(x) + g(x) dx = f(x) dx + g(x) dx Z Z cf(x) dx = c f(x) dx.

14 R 2 √ 2 Example 9.9. Evaluate 3x − 5 x + x2 dx. Theorem 9.10 (Basic Trig integrals). Z Z sin x dx = − cos x + C cos x dx = sin x + C.

Theorem 9.11 (Antiderivative of ex). Z ex dx = ex + C.