Math 131. Increasing/Decreasing Functions and First Derivative Test Larson Section 3.3

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1 Math 131. Increasing/Decreasing Functions and First Derivative Test Larson Section 3.3 Increasing and Decreasing Functions. A function f is increasing on an interval if for any two numbers x 1 and x 2 in the interval, x 1 < x 2 implies f(x 1 ) < f(x 2 ). A function f is decreasing on an interval if for any two numbers x 1 and x 2 in the interval, x 1 < x 2 implies f(x 1 ) > f(x 2 ). A function is increasing if as x moves to the right, the graph moves upward, and decreasing if the graph moves downward. The Mean Value theorem allows us to test for increasing and decreasing functions using the first derivative. Test for Increasing and Decreasing Functions. Let f be continuous on the closed interval [a, b] and differentiable in the open interval (a, b). 1. If f (x) > 0 for all x in (a, b), then f is increasing on [a, b]. 2. If f (x) < 0 for all x in (a, b), then f is decreasing on [a, b]. 3. If f (x) = 0 for all x in (a, b), then f is constant on [a, b]. Proof. We will prove the second case; the first case is done in your text. Suppose f (x) < 0 for all x (a, b). Now let x 1 < x 2 be any points in the interval [a, b]. By the Mean Value theorem, there is a number c between x 1 and x 2 with f (c) = f(x 2) f(x 1 ) x 2 x 1 This means f(x 2 ) f(x 1 ) = f (c)(x 2 x 1 ) < 0 since f (x) < 0. Therefore, f(x 2 ) < f(x 1 ) if x 1 < x 2 in the interval, and this means f is decreasing on the interval [a, b]. To determine where a function is increasing or decreasing we usually: Find the critical numbers of the function, and use them to separate the number line or domain of the function into intervals. Use a sign analysis on the derivative, or test points with the derivative to determine whether the derivative is positive or negative on the interval. Apply the previous theorem to determine whether the function is increasing or decreasing. Example 1. Let f(x) = x x 2 135x 2. (a) Find the critical numbers of f. (b) Find open intervals on which f is increasing. (c) Find open intervals on which f is decreasing.

2 Solution: (a) Because f(x) = x x 2 135x 2 is a polynomial, it is differentiable everywhere, so the critical numbers of f are the values of x where f (x) = 0. Now, f (x) = 3x x 135 = 3(x 2 14x + 45) = 3(x 5)(x 9) Thus f (x) = 0 when x = 5 and x = 9, and so these are the critical numbers of f. (b) If 5 < x < 9, then f (x) = 3(pos num)(neg num) > 0 and so f is increasing on (5, 9). (c) If x < 5, then and if x > 9, then f (x) = 3(neg num)(neg num) < 0 f (x) = 3(pos num)(pos num) < 0. Thus f is decreasing on the intervals (, 5) and (9, ). The following test lets you determine whether a function has a maximum, minimum or neither at a critical number. The First Derivative Test. Suppose f is continuous on an open interval I containing c, and suppose c is a critical number of f. If f is differentiable on the interval, except possibly at c, then f(c) can be classified as follows. 1. If f (x) changes from negative to positive at c, then f has a relative minimum at (c, f(c)). 2. If f (x) changes from positive to negative at c, then f has a relative maximum at (c, f(c)). 3. If f (x) is positive on both sides of c, or negative on both sides of c, then f(c) is neither a relative minimum nor relative maximum. The following diagram illustrates why this theorem is true, the proof essentially puts the diagram to words using the theorem on increasing and decreasing functions.

3 Example 2. Let f(x) = 2x. Find open intervals on which f is increasing, and open x intervals on which f is decreasing. Then classify the relative extrema of f. Solution: First, we use the quotient rule to find f ( ) (1)(x f 2 + 1) (x)(2x) (x) = 2 (x 2 + 1) 2 ( ) 1 x 2 = 2 (x 2 + 1) 2 Then f (x) > 0 if 1 < x < 1, and f (x) < 0 if x < 1 or x > 1. Thus f is decreasing on the open intervals (, 1) and on (1, ). The function f is increasing on the open interval ( 1, 1). The critical numbers of f are x = 1, and x = 1. The function f has a relative minimum at x = 1 and a relative maximum at x = 1. Example 3. Let f(x) = x 3 + 9x x 2. (a) Find the critical numbers of f. (b) Find open intervals on which f is increasing. (c) Find open intervals on which f is decreasing. (d) Classify the relative extrema of f as relative maximums, relative minimums or neither. Solution: We will first find the critical numbers of f. Because f(x) = x 3 + 9x x 2 is a polynomial, it is differentiable everywhere, so the critical numbers of f are the values of x where f (x) = 0. Now, f (x) = 3x x + 15 = 3(x 2 + 6x + 5) = 3(x + 5)(x + 1) (a) Thus f (x) = 0 when x = 5 and x = 1, and so these are the critical numbers of f. (b) If x < 5, then f (x) = 3(neg num)(neg num) > 0 and if x > 1, then f (x) = 3(pos num)(pos num) > 0. Thus f is increasing on the intervals (, 5) and ( 1, ). (c) If 5 < x < 1, then f (x) = 3(pos num)(neg num) < 0

4 and so f is decreasing on ( 5, 1). (d) f has a relative minimum at x = 1, because f changes from negative to positive at x = 1, and f has a relative maximum at x = 5 because f changes from positive to negative at x = 5. Example 4. A particle moves along a straight line and its position at time t is given by s(t) = t3 3 3t2 + 8t + 3 where s is measured in feet and t in seconds. (a) Find the velocity v(t) of the particle at time t 0. (b) Identify time intervals when the particle is moving in a positive direction. (c) Identifty time intervals when the particle is moving in a negative direction. (d) Identify the times when the particle changes direction. Solution: (a) v(t) = s (t) = t 2 2(3)t + 8 = t 2 6t + 8. (b) and (c): First, we solve v(t) = 0, that is t 2 6t + 8 = 0 and so t 2 6t + 8 = (t 2)(t 4) = 0. The intervals to consider are the ones separated by 2 and 4: (0, 2), (2, 4) and (4, ). Using test values in the intervals, you will find v(t) > 0 on (0, 2) and (4, ) (for example, to test the interval (4, ), pick any value in it, say 5, and compute v(5) = 5 2 (6)(5)+8 = 3 > 0 and so v(t) > 0 on the entire interval (4, ); you can test the other two intervals similarly). You will find v(t) < 0 on (2, 4). [The sign of v(t) can also be done by sign analysis from the factored form of v(t).] (b) The particle is moving in a positive direction on the intervals (0, 2) and (4, ). (c) The particle is moving in a negative direction on the interval (2, 4). (d) The particle changes direction when t = 2 and t = 4. Example 5. Let f(x) = x2 2x + 1 x + 1 (a) Find the critical numbers of f. (b) Find open intervals where f is increasing and and where is f decreasing.

5 (c) Classify the extrema of f as local maxima and local minima. Solution: Compute f (x) = (2x 2)(x + 1) (x 1)2 (x + 1) 2 = (x 1)(x + 3) (x + 1) 2 (a) The critical numbers for f are x = 1 and x = 3. Note that f has a vertical asymptote x = 1. (b) f is increasing when x < 3, x > 1. 1 < x < 1. f is decreasing when 3 < x < 1 and (c) f has a local maximum ( 3, 8) and a local minimum (1, 0) A graph of f is given below for reference Example 6. The graphs of the derivatives of f(x) and g(x) are given below. Graph of f (x) y x Graph of g (x) y x (a) Use the graph of f (x) to (i) determine the critical numbers of f; (ii) find open intervals where f is increasing; (iii) find open intervals where f is decreasing; and (iv) classify the relative extrema of f.

6 (b) Use the graph of g (x) to (i) determine the critical numbers of g; (ii) find open intervals where g is increasing; (iii) find open intervals where g is decreasing; and (iv) classify the relative extrema of g. Solution: (a) The critical number of f is x = 1, because f ( 1) = 0. The function f is increasing on the interval (, 1) because f (x) > 0 on that interval since the graph of f (x) is above the x-axis there. The function f is decreasing on the interval ( 1, ) because f (x) < 0 on that interval since the graph of f (x) is below the x-axis there. The function f has a relative maximum at x = 1 because f (x) changes from positive to negative at 1. (b) The critical numbers of g are x = 4 and x = 0 because those are the values where g (x) = 0 (the graph of g (x) crosses the x-axis at those two values). The function g is increasing on the open intervals (, 4) and (0, ) because g (x) > 0 there (this is where the graph of g (x) is above the x-axis). The function g is decreasing on the open interval ( 4, 0) because g (x) < 0 there (this is where the graph of g (x) is below the x-axis). The function g has a relative minimum at x = 0 because g (x) changes from negative to positive at x = 0; and the function g has a relative maximum at x = 4 because g (x) changes from positive to negative at x = 4.