Mark M. answered 04/27/24
Tutor
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f(x) = x5/4(x - 4)2
f'(x) = (5/4)x1/4(x - 4)2 + x5/4(2)(x - 4) = x1/4(x - 4)[(5/4)(x - 4) + 2x] = x1/4(x - 4)[(5/4)x - 5 + 2x]
f'(x) = x1/4(x -4)[(13/4)x - 5] = 0 when x = 0, 4, 20/13
f has critical points when x = 0, 4, 20/13