5.2 The Derivative Function 191 Where f (x) = 0 the graph of f has a horizontal tangent line. 2x + 3 = 0 2x =−3 x=−1.5 Below we graph f and f . Notice that the answers make sense in terms of our intuitive ideas about slope. f ' f '(x)=2x+3 xx f f(x)= x +3x 2 –3 (–1.5, –2.25) –1.5 Figure 5.15 ◆ EXERCISE 5.2 Use the limit definition of derivative to find the derivatives of x 2 and 3x. Conclude that the derivative of x 2 + 3x is the sum of the derivatives of x 2 and 3x. Notice that so far the strategy that has been working for computing derivatives is to simplify the difference quotient to the point that we can factor an h from the numerator to cancel with the h in the denominator (assuming that h = 0) and, once done, the limit becomes apparent. Sometimes the simplification allowing cancellation is a bit more com- plicated, as illustrated in the next problem. You should compute some simpler derivatives on your own before reading the next example. It is designed for your second reading of the chapter. ◆ EXAMPLE 5.5 Let f(x)= √ x. Find f (x). SOLUTION f (x) = lim h→0 f(x +h) − f(x) h = lim h→0 √ x + h − √ x h We have a dilemma here; to make progress we must cancel the h in the denominator. This means getting rid of the square roots. We’re working with an expression, so our options are to multiply by 1 (multiply numerator and denominator by the same nonzero quantity) or to add zero. The former will be most productive. Multiplying the expression √ A − √ B by ( √ A + √ B) will eliminate the square roots. 8 ( √ A − √ B) is called the conjugate of ( √ A + √ B). ( √ A − √ B)( √ A + √ B) =A − √ AB + √ AB − B = A − B 8 Multiplying ( √ A + √ B) by itself does not eliminate the square roots. ( √ A + √ B)( √ A + √ B) =A+2 √ AB +B, which leaves the square root in the mixed term. 192 CHAPTER 5 The Derivative Function This algebraic maneuver is worth stashing away in your mind for future reference. f (x) = lim h→0 √ x + h − √ x h Multiply numerator and denominator by √ x + h + √ x. = lim h→0 ( √ x + h − √ x) h · ( √ x + h + √ x) ( √ x + h + √ x) = lim h→0 x + h − x h( √ x + h + √ x) = lim h→0 h h( √ x + h + √ x) Since h = 0, h h = 1. = lim h→0 1 √ x + h + √ x = 1 √ x + √ x = 1 2 √ x So the derivative of x 1/2 is 1 2 x −1/2 . ◆ Answers to Selected Exercises Answers to Exercise 5.1 (a) f (x) = lim h→0 f(x +h) − f(x) h = lim h→0 m(x + h) + b − (mx + b) h = lim h→0 mx + mh + b − mx − b h = lim h→0 mh h = m (b) k =mx +k where m = 0, so the derivative of k is zero. PROBLEMS FOR SECTION 5.2 1. Use the limit definition of derivative to show that the derivative of the linear function f(x)=ax +b is a. Why is this exactly what you would expect? You have shown that the derivative of a constant is zero. Explain, and explain why this is exactly what you would expect. 5.2 The Derivative Function 193 2. Use the limit definition of derivative to find the derivative of f(x)=kx 2 . 3. Let f(x)=x 2 .Find the point at which the line tangent to f(x)at x = 2 intersects the line tangent to f(x)at x =−1. 4. Use the limit definition of derivative to find the derivative of f(x)=x 3 . 5. Using the limit definition of the derivative, find f (x) if f(x)=(x − 1) 2 . 6. Let g(x) = x 2x+5 . Using the limit definition of derivative, find g (x). For Problems 7 through 13, find f (x), f (0), f (2), and f (−1). 7. f(x)=3x+5 8. f(x)=πx − √ 3 9. f(x)= 2x−5 3 10. f(x)=π(x +7) −2 11. f(x)=x 2 12. f(x)= 1 x 13. f(x)= x+π 2 14. Suppose f(x)=x 2 . For what value(s) of x is the instantaneous rate of change of f at x equal to the average rate of change of f on the specified interval? Illustrate your answers with graphs. (a) the interval [0, 3] (b) the interval [1, 4] 15. Let g(x) = 1 x . For what value(s) of x is the slope of the tangent line to g equal to the average rate of change of g on the interval indicated? Illustrate your answer to parts (a) and (b) with pictures. (a) [ 1 2 , 2] (b) [1, 4] (c) [c, d] d>c>0 16. Let f(x)= 1 x 2 . Let P and Q be points on the graph of f with coordinates (x, f(x)) and (x + x, f(x+x)), respectively. (a) Find the slope of the secant line through P and Q. Simplify your answer as much as possible. (b) By calculating the appropriate limit, find the slope of the tangent line to f(x) at point P . 17. We showed that the derivative of √ x (or x 1 2 )is 1 2 1 √ x (or 1 2 x − 1 2 ). Here we focus on f(x)= √ x − 1. 194 CHAPTER 5 The Derivative Function (a) How is the graph of √ x − 1 related to that of √ x? (b) How is the graph of the derivative of √ x − 1 related to that of the derivative of √ x? Illustrate with a rough sketch. (c) Given your answer to part (b) explain why d dx √ x x=4 = d dx √ x − 1 x=5 . In other words, explain why the derivative of √ x at x = 4 is equal to the derivative of √ x − 1 evaluated at x = 5. (d) Show that f (5) = 1 4 using the limit definition of derivative: f (5) = lim x→5 f(x)−f(5) x − 5 . (You’ll need to rationalize the numerator.) 18. Show that d dx √ x + 8 = 1 2 √ x+8 using the limit definition of derivative. You’ll use different versions of the definition in parts (a) and (b). In both cases it will be necessary to rationalize the numerator in order to evaluate the limit. (a) f (x) = lim h→0 f(x+h)−f(x) h (b) f (x) = lim b→x f(b)−f(x) b−x 19. Let f(x)=x − 1 2 . Use the limit definition of derivative to show that f (x) =− 1 2 x − 3 2 . 5.3 QUALITATIVE INTERPRETATION OF THE DERIVATIVE Key Notions In this section and the following, we’ll interpret the derivative function f as the slope function. Assertions will be made that should make sense intuitively. However, we will not prove these assertions and “facts” here; proofs will be delayed until after Chapter 7. Fact: Where f is increasing, the slope of its graph is nonnegative; where f is decreasing, its slope is nonpositive. 9 9 You may wonder why we say that f is increasing implies that the slope is nonnegative rather than just saying that the slope is positive. Consider a function like f(x)=x 3 . It isincreasing everywhere, yet its slope at x = 0 is zero; locally it looks like a horizontal line around (0, 0). y y=x 3 xx y magnified picture 5.3 Qualitative Interpretation of the Derivative 195 Where f is increasing, f ≥ 0; where f is decreasing, f ≤ 0. It follows that where f is positive, the graph of f is increasing (from left to right); where f is negative, the graph of f is decreasing (from left to right); where f is zero, the graph of f locally looks like a horizontal line. Concavity is determined by whether f is increasing or decreasing. Where f > 0 and f is increasing, the graph of f looks like Figure 5.16(a). Where f > 0 and f is decreasing, the graph of f looks like Figure 5.16(b): Where f < 0 and f is increasing, the graph of f looks like Figure 5.16(c): Where f < 0 and f is decreasing, the graph of f looks like Figure 5.16(d): 10 (a) (b) (c) (d) Figure 5.16 Where the slope of f is increasing, we say the graph of f is concave up. In other words, where f is increasing the graph of f is concave up. Figures 5.16(a) and (c) are examples of concave-up graphs. concave up Figure 5.17 Where the slope of f is decreasing, we say the graph of f is concave down. In other words, where f is decreasing, the graph of f is concave down. Figures 5.16(b) and (d) are examples of concave-down graphs. 11 concave down Figure 5.18 10 If f < 0 and f is decreasing, then f is becoming increasingly negative as the independent variable increases. 11 Concave up “holds water”; concave down “spills water.” 196 CHAPTER 5 The Derivative Function COMMON ERROR Consider the graph in Figure 5.16(c). A common error is to think that the slope is decreasing, because as x increases the drop is becoming more gentle. This is incorrect. As the slope changes from, say, −2to−1, the slope is increasing (becoming less negative); similarly, if the temperature goes from −10 degrees to −5 degrees, we say that the temperature is increasing. As the slope changes from −2to−1, the steepness of the line decreases, but the slope of the line increases. Should you ever become confused, label the slopes at a few points and put these slopes in order on a number line. Interpreting the Derivative Function Graphically Let’s look qualitatively at the relationship between a function and its derivative function. Sketching the Derivative Function Given the Graph of f The problem of sketching f when we are given the graph of f is equivalent to the problem of sketching the velocity graph for a trip when we know the graph of position versus time. When sketching the derivative function, begin with the most fundamental questions: Where is f zero? undefined? positive? negative? Note that, like any other function, f cannot change sign without passing through a point at which it is either zero or undefined. COMMON ERROR A standard mistake of the novice derivative sketcher is to get mesmerized by an internal conversation about where the slope is increasing and decreasing without first considering the question of whether the slope is positive or negative. Keep your priorities in order! Observations about the concavity of f are fine tuning and are not the first order of business. On the following page are several worked examples. Where f has a horizontal tangent line, f is zero. Where f is discontinuous or has a sharp corner, f is undefined. We mark the values where f is zero or undefined on a number line. At the bottom we have tracked the sign of f ; above the number line we have used arrows to indicate where the graph of f is increasing and decreasing. Where f is increasing, f ≥ 0; the region above the x-axis on the graph of f is shaded to indicate this. Where f is decreasing, f ≤ 0; the region below the x-axis on the graph of f is shaded. The graph of f must lie in the shaded regions. 5.3 Qualitative Interpretation of the Derivative 197 ◆ EXAMPLE 5.6 Given the graph of f , produce a rough sketch of f . abcd f x abc x d _ 0 + 0 _ 0 + 0 sign of f ' graph of f f ' x rough sketch of f ' f '=0 at x=a, b, c, d a b c d ◆ ◆ EXAMPLE 5.7 Given the graph of f , produce a rough sketch of f . a x f __ sign of f ' graph of f a x undefined a x f ' rough sketch of f ' f ' undefined at x=a ◆ 198 CHAPTER 5 The Derivative Function ◆ EXAMPLE 5.8 Given the graph of f , produce a rough graph of f . f abc x a x bc _ 0+ 0 graph of f undefined _ sign of f ' + f ' x rough sketch of f ' f '=0 at x=a, c f 'undefined at x=0=b ◆ Getting Information About f Given the Graph of f The problem of sketching f given the graph of f is equivalent to the problem of sketching the position versus time graph for a trip given the graph of velocity as a function of time. Unless we know either where the trip began or the location at some particular time, we cannot produce the actual graph of the trip; we can only get the general shape of the trip graph without information as to position at any time. For instance, if we are told that a cyclist travels 12 miles per hour for 2 hours, we know that he has traveled 24 miles, but we have no idea at all as to where he started or finished his trip. If we are given his position at some particular time, then we can piece together a picture of his position throughout the trip. Analogously, because f gives information only about the slope of f , we can simply obtain information about the shape of f , not its vertical position. To make a rough sketch of the shape of f given f , begin again with the most important features of f : Where is f zero? undefined? positive? negative? Where f is positive, the graph of f is increasing; where f is negative, the graph of f is decreasing; where f is zero, the graph of f locally looks like a horizontal line. Keep clear in your mind the difference between f being positive and being increasing. When you look at a graph of f it is easy to get swept away by the shape of the graph. Hold back! Sometimes the graph of f gives you more information than you can easily process. We suggest organizing the sign information on a number line. Begin by asking yourself (in a calm voice) “Where is f zero? Where is f positive?” 5.3 Qualitative Interpretation of the Derivative 199 As pointed out in Chapter 2, sign information is a fundamental characteristic of a function; we see now that the sign of f gives vital information. Begin by constructing a number line highlighting this information. This method is illustrated in the following example. ◆ EXAMPLE 5.9 Use the graph of f to answer the following questions. i. Where on the interval [−3, 10] is the value of f the largest? ii. Which is larger, f(2)or f(5)? iii. Where on the interval [−3, 10] is f increasing most rapidly? iv. Where on the interval [−3, 10] is the graph of f concave up? –3 –2 –11023 4 56 7 8 9 10 11 x f ' (NOT f) Figure 5.19 SOLUTION i. Begin by constructing a number line with information about the sign of f and its implications for the graph of f . –2 _ 0 + sign of f ' graph of f 7 0 _ x Figure 5.20 From this analysis we see that f is largest at either x =−3orx=7. Since f is positive on [−2, 7] the graph of f is increasing on the entire interval. The increase on the interval (−2, 7) is substantially greater than the decrease on (−3, −2) so f(7)>f(−3). f is largest at x = 7. ii. f(5)>f(2)since f is increasing on [−2, 7]. iii. f is increasing most rapidly where f is greatest. This is at x = 3. 200 CHAPTER 5 The Derivative Function iv. f is concave up where f is increasing. This is on the intervals [−3, 3], [5, 6], and [9, 10]. ◆ ◆ EXAMPLE 5.10 For each of the following, given the graph of f , sketch possible graphs of f . Because f gives us only information about the slope of f , there are infinitely many choices for f ; each is a vertical translate of graphs given. The most constructive way to “read” this example is to do so actively. Do the problems on your own and then look at the solutions. f ' f ' f ' xxx 1 2 (a)(b)(c) x x f ' f ' –1 1 (d) (e) SOLUTION x f ' x f graph of f sign of f ' + 2 (a) x f ' f 1 x graph of f sign of f ' + 0 _ 1 1 (b) f ' f xx graph of f sign of f ' + (c) . = 1 x . For what value(s) of x is the slope of the tangent line to g equal to the average rate of change of g on the interval indicated? Illustrate your answer to parts (a) and (b) with pictures. (a). denominator. This means getting rid of the square roots. We’re working with an expression, so our options are to multiply by 1 (multiply numerator and denominator by the same nonzero quantity) or to. f(x)=x 2 . For what value(s) of x is the instantaneous rate of change of f at x equal to the average rate of change of f on the specified interval? Illustrate your answers with graphs. (a) the