What are the local maxima and minima of #f(x) = 4x^3 + 3x^2 - 6x + 1#?

Question

Cameron Alexander · Accepted Answer

Polynomials are differentiable everywhere , so look for the critical values by simply finding the solutions to #f'=0#

#f'=12x^2+6x-6=0#

Using algebra to solve this simple quadratic equation:

#x=-1# and #x=1/2#

Determine if these are min or max by plugging into the second derivative:

#f'' = 24x+6#

#f''(-1) <0#, so -1 is a maximum

#f''(1/2)>0#, so 1/2 is a minimum

hope that helped

Emily Ammerman · Answer

undefined To find the local maxima and minima of $ f(x) = 4x^3 + 3x^2 - 6x + 1 $, first, find its derivative $ f'(x) $. Then, set $ f'(x) = 0 $ and solve for $ x $. The values of $ x $ obtained will correspond to critical points. Test these critical points using the second derivative test to determine whether they correspond to local maxima, minima, or points of inflection.

$ f'(x) = 12x^2 + 6x - 6 $

Set $ f'(x) = 0 $ and solve for $ x $:

$ 12x^2 + 6x - 6 = 0 $

Solving this quadratic equation yields two solutions: $ x = -1 $ and $ x = \frac{1}{2} $.

To classify these critical points, find $ f''(x) $, the second derivative:

$ f''(x) = 24x + 6 $

Evaluate $ f''(-1) $ and $ f''\left(\frac{1}{2}\right) $:

$ f''(-1) = 24(-1) + 6 = -18 $ (negative, indicating concave down)

$ f''\left(\frac{1}{2}\right) = 24\left(\frac{1}{2}\right) + 6 = 18 $ (positive, indicating concave up)

Since $ f''(-1) $ is negative, $ x = -1 $ corresponds to a local maximum, and since $ f''\left(\frac{1}{2}\right) $ is positive, $ x = \frac{1}{2} $ corresponds to a local minimum.

So, the local maximum is at $ x = -1 $ and the local minimum is at $ x = \frac{1}{2} $.