Find all absolute and relative extreme values of the function f(x)=−x3+2x2−x−1, on the interval [-2,2].

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Ruth Phillips · Accepted Answer

Step 1The given function is f(x)=−x3+2x2−x−1 on [-2,2].Compute the critical points as follows.f&#039;(x)=0ddx(−x3+2x2−x−1)=0−3x2+4x−1=0−3x2+3x+x−1=0-3x(x-1)+(x-1)=0(x-1)(-3x+1)=0x=1,13Thus, the critical points are x=1, 13.Step 2Obtain the second derivative as follows.f″(x)=ddx(−3x2+4x−1)=-6x+4Note that at x=1, f&#039;&#039;(x) &amp;lt; 0. The relative maximum exist at x=1.at x=13,f″(x)&amp;gt;0. The relative minimum exist at x=13.Compute the function value at critical values and end points of [-2,2] as follows. Step 3At x=−2,f(−2)=−(−2)3+2(−2)2−(−2)−1=17At x=13,f(13)=−(13)3+2(13)2−(13)−1=−3127At x=1,f(1)=−(1)3+2(1)2−(1)−1=−1At x=2,f(2)=−(2)3+2(2)2−(2)−1=−3Thus,Absolute maximum value is 17.Absolute minimum value is -3.Relative maximum is -1Relative minimum is −3127.

Find all absolute and relative extreme values of the function

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