Explanation Definition used: First derivative test: Suppose c to be a critical number for a function f . Assume that f is continuous on ( a , b ) and differentiable on ( a , b ) except at c . The interval ( a , b ) should have only one critical number of f . Relative maximum : A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) > 0 in the interval ( a , c ) and f ′ ( x ) < 0 in the interval ( c , b ) . Relative minimum: A function is said to have a relative maximum f ( c ) at c if f ′ ( x ) < 0 in the interval ( a , c ) and f ′ ( x ) > 0 in the interval ( c , b ) . Calculation: The given function is f ( x ) = x 2 − 6 x + 9 x + 2 . Obtain the derivative of the function as follows. f ′ ( x ) = ( x 2 − 6 x + 9 x + 2 ) ′ = ( x + 2 ) ( 2 x − 6 ) − ( x 2 − 6 x + 9 ) ( x + 2 ) 2 = 2 x 2 − 6 x + 4 x − 12 − ( x 2 − 6 x + 9 ) ( x + 2 ) 2 = 2 x 2 − 2 x − 12 − x 2 + 6 x − 9 ( x + 2 ) 2 Simplify further as follows. f ′ ( x ) = 2 x 2 − 2 x − 12 − x 2 + 6 x − 9 ( x + 2 ) 2 = x 2 + 4 x − 21 ( x + 2 ) 2 Set f ′ ( c ) = 0 and find the value of c . c 2 + 4 c − 21 ( c + 2 ) 2 = 0 c 2 + 4 c − 21 = 0 c 2 + 7 c − 3 c − 21 = 0 c ( c + 7 ) − 3 ( c + 7 ) = 0 Simplify further as follows. c ( c + 7 ) − 3 ( c + 7 ) = 0 ( c + 7 ) ( c − 3 ) = 0 c + 7 = 0 or c − 3 = 0 c = − 7 or c = 3 Here, the derivative does not exist at a c = − 2 and hence it cannot be a critical number. Therefore, the critical numbers are c = − 7 and c = 3 . These points determine the intervals ( − ∞ , − 7 ) , ( − 7 , 3 ) and ( 3 , ∞ )

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Chapter 13.2, Problem 28E

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To find: The x-values of all points where the function f(x)=x2−6x+9x+2 has relative extrema and the value of relative extrema.

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Chapter 13.2, Problem 27E

Chapter 13.2, Problem 29E

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(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

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EBK FINITE MATHEMATICS AND CALCULUS WIT

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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The x -values of all points where the function f ( x ) = x 2 − 6 x + 9 x + 2 has relative extrema and the value of relative extrema.

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To find: The x-values of all points where the function f(x)=x2−6x+9x+2 has relative extrema and the value of relative extrema.

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