Find the local maximum and minimum values of using both the First and Second Derivative Tests. Which method do you prefer? f(x)=1+3x2−2x3

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Antum1978 · Accepted Answer

Step 1f(x)=1+3x2−2x3Step 2Find where f′=00=6x−6x2=6x(1−x)x=0,1Step 31st Derivative TestTest intervals based on the critical numbers:(−∞,0): f′(−1)=−12&amp;lt;0⇒f is decreasing(0,1): f′(0.5)=1.5&amp;gt;0⇒ f is increasing(1,∞): f′(2)=−12&amp;lt;0⇒f is decreasingFrom this we can conclude:Local minimum at (0,f(0))=(0,1)Local maximum at (1,f(1))=(1,2)

Warajected53 · Accepted Answer

Step 4Find f&quot;f(x)=6−12xStep 52nd Derivative TestTest the points where f′=0∈ff(0)=6&amp;gt;0. Concave up at this point, so x=0 is a local min.f(1)=−6&amp;lt;0. Concave down at this point, so x=1 is a local max.Step 6The 2nd Derivative Test is preferable because it requires less testing, though sometimes the time savings may be offset by the effort needed to get the 2nd Derivative if the function is complicated. local min: (0,2)         local max: (1,2)

Find the local maximum and minimum values of using both the First and Second Der

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