Does the function #f(x)=-8x +x^2# have a maximum or minimum?

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Answer 1 · 2017-08-03T17:07:41

There is a minimum at #(4,-16)#

We need

#(x^n)'=nx^(n-1)#

Our function is

#f(x)=x^2-8x#

We calculate the first derivative,

#f'(x)=2x-8#

The critical point is when #f'(x)=0#

#2x-8=0#, #=>#, #x=4#

The second derivative is

#f''(x)=2#

As

#f''(x)>0#, the function is convex at #x=4#

graph{x^2-8x [-31.13, 33.82, -22.35, 10.13]}