Graphs and the Derivative Chapter 5 Graphs and the Derivative

Increasing and Decreasing Functions Section 5.1 Increasing and Decreasing Functions

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Your Turn 1 Find where the function is increasing and decreasing. Solution: Moving from left to right, the function is decreasing for x-values up to −1, then increasing for x-values from to −1 to 2. For x-values from 2 to 4, decreasing and increasing for all x-values larger than 4. In interval notation, the function is increasing on (−1, 2) and (4, ∞). Function is decreasing on (−∞, −1) and (2,4).

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Your Turn 2 Find the intervals in which is increasing or decreasing. Graph the function. Solution: Here To find the critical numbers, set this derivative equal to 0 and solve the resulting equation by factoring. Continued

Your Turn 2 Continued The tangent line is horizontal at −3 or 5/3. Since there are no values of x where derivative fails to exist, the only critical numbers are −3 and 5/3. To determine where the function is increasing or decreasing, locate −3 and 5/3 on a number line. The number line is divided into three intervals namely, (−∞, −3), ( −3, 5/3), and (5/3, ∞). Now choose any value of x in the interval (−∞, −3), choosing − 4 and evaluating which is negative. Therefore, f is decreasing on (−∞, −3). Continued

Your Turn 2 Continued Now choose any value of x in the interval ( −3, 5/3), choosing x = 0 and evaluating which is positive. Therefore, f is increasing on ( −3, 5/3). Now choose any value of x in the interval (5/3, ∞), choosing x = 2 and evaluating which is negative. Therefore, f is decreasing on (5/3, ∞). Continued

Your Turn 2 Continued

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Section 5.2 Relative Extrema

Your Turn 1 Identify the x-values of all points where the graph has relative extrema. Solution:

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Your Turn 2 Find all relative extrema Solution: Here To find the critical numbers, set this derivative equal to 0 and solve the resulting equation by factoring. Continued

Your Turn 2 Continued The tangent line is horizontal at −3 or 5/3. Since there are no values of x where derivative fails to exist, the only critical numbers are −3 and 5/3. The number line is divided into three intervals namely, (−∞, −3), ( −3, 5/3), and (5/3, ∞). Any number from each of the three intervals can be used as a test point to find the sign of in each interval. Using − 4, 0, and 2 gives the following information. Continued

Your Turn 2 Continued Thus derivative is negative on(−∞, −3), positive on ( −3, 5/3), and negative on (5/3, ∞). Using the first derivative test, this means that the function has a relative maximum of f (5/3) = 670/27 and f has a relative minimum of f (−3) = −26.

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Higher Derivatives, Concavity, and the Second Derivative Test Section 5.3 Higher Derivatives, Concavity, and the Second Derivative Test

Your Turn 1 Solution: To find the second derivative of f (x), find the first derivative, and then take its derivative.

Your Turn 2(a) Solution : Here, using the chain rule,

Your Turn 2(b) Solution: Use the product rule,

Your Turn 2(c) Solution: Here, we need the quotient rule.

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Your Turn 5 Solution: First, find the points where the derivative is 0. Here Now use the second derivative test. The second derivative is Continued

Your Turn 5 Continued so that by second derivative test, −3 leads to a relative minimum of Also, when x = 4, so that by second derivative test, 4 leads to a relative maximum

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Section 5.4 Curve Sketching

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