Find the values of x where the extreme values of the function y=x^3-7x-5x

Okay well then we can simplify it first by collecting like term: y=(x^3)-7x-5x y=x^3-12x then we find the derivative y'=3x^2-12 next we set the derivative to zero: 3x^2-12=0-->3(x^2-4)=0-->3(x-2)(x+2)=0 Thus our critical number are: x-2=0--->x=2 and x+2=0--->x=-2 Which is of course makes no sense since this not one of your possible choicesSHOW FULL ANSWER

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