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Q. 39

Found in: Page 261


Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

Use the first derivative test to determine the local extrema of each function in Exercises 39- 50. Then verify your algebraic answers with graphs from a calculator or graphing utility.


Ans: The local extrema value of f(x) are x=0,4

See the step by step solution

Step by Step Solution

Step 1. Given Information


Step 2. Finding the derivative of the function

Rewriting the function by simplifying it,

f(x)=(x-2)2(x+1)f(x)=(x2-4x+4)(1+x)f(x)=x2+x3-4x-4x2+4+4xf(x)=x3-3x2+4f'(x)=3x2-6xlets, f'(x)=03x2-6x=0 3x(x-2)=0x={0,2}substitute x=-1,1,3 into f'(x)=3x2-6xf'(-1)=3(-1)2-6(-1) =9>0f'(1)=3(1)2-6(1) =-3<0f'(3)=3(3)2-6(3) =9>0

Step 3. Finding local extrema of the function on the number line(Sign chart):

From the graph identify the local minimum at x=2 indicated on the chartwith the first derivative test, since f'(x) changes sign from negative to positive at x=2.f(0)=(0-2)2(1+0) =4 (local maximum)f(2)=(2-2)2(1+2) =0 (local minimum)the extreme values of the function are x={0,4}

Step 4. Verifying algebraic answers with graphs :


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