The critical values of the function are x = 0 and x = 64
The critical values of a function f(x) are the values of x for which f'(x) = 0.
Given,
f(x) = [tex]\frac{x^{4} }{5(x-8)^{2} }[/tex]
The derivative is found as follows, applying the quotient rule:
f'(x) = [tex]\frac{[5(x^{4} )(x-8)^{2}-5x^{4} [(x-8)^{2} ]' }{[5(x-8)^{2} ]^{2} }[/tex]
= [tex]\frac{2x^{3}(x-64) }{5(x-8)^{3} }[/tex]
The zeros of the function are the zeros of the numerator, thus:
[tex]2x^{3} (x-64) =0\\2x^{3} =0-------- > x =0\\[/tex]
x - 64 = 0------------->x = 64
The critical values are x = 0 and x = 64
Learn more about critical values here:https://brainly.com/question/12958088
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