What are the local extrema, if any, of #f (x) = x^3-12x+2 #?

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Answer 1 · 2015-11-09T16:55:16

The function has 2 extrema:

#f_{max}(-2)=18# and #f_{min}(2)=-14#

We have a function: #f(x)=x^3-12x+2#

To find extrema we calculate derivative

#f'(x)=3x^2-12#

The first condition to find extreme points is that such points exist only where #f'(x)=0#

#3x^2-12=0#

#3(x^2-4)=0)#

#3(x-2)(x+2)=0#

#x=2 vv x=-2#

Now we have to check if the derivative changes sign at the calcolated points:

graph{x^2-4 [-10, 10, -4.96, 13.06]}

From the graph we can see that #f(x)# has maximum for #x=-2# and minimum for #x=2#.

Final step is to calculate the values #f(-2)# and #f(2)#