Question

Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.

p(t) = 2t³ + 3t² - 36t

Accepted Answer

First, find the first derivative of the function $$ p(t) = 2t^3 + 3t^2 - 36t . $$This involves using the power rule for differentiation. The first derivative $$ p'(t)  is $$calculated as follows: $$ p'(t) = \frac{d}{dt}(2t^3) + \frac{d}{dt}(3t^2) - \frac{d}{dt}(36t) . $$Simplify the expression for the first derivative: $$ p'(t) = 6t^2 + 6t - 36 . To $$find the critical points, set the first derivative equal to zero and solve for $$ t $$: $$ 6t^2 + 6t - 36 = 0 . $$This is a quadratic equation, which can be solved using the quadratic formula $$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$, where $$ a = 6 $$, $$ b = 6 $$, and $$ c = -36 . $$Next, find the second derivative of the function $$ p(t)  to $$apply the Second Derivative Test. Differentiate $$ p'(t) = 6t^2 + 6t - 36  to $$get $$ p''(t) = 12t + 6 . $$Evaluate the second derivative at each critical point found in step 3. If $$ p''(t) \u003e 0 $$, the function has a local minimum at that point. If $$ p''(t) \u003c 0 $$, the function has a local maximum. If $$ p''(t) = 0 $$, the test is inconclusive.

Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.

p(t) = 2t³ + 3t² - 36t

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Key Concepts

Critical Points