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5 votes
F(x)=x^3+5/2x^2-28x
Find critical numbers

User MikeyE
by
8.5k points

1 Answer

5 votes

Answer:

x = 0.761, 2, 14

Explanation:

critical numbers are points where f'(x) = 0 or undefined and the endpoints

since there are no endpoints for the domain, we just consider the points where f'(x) is 0 or undefined


(d)/(dx)
((x^3+5)/(2x^2-28x)) =
((3x^2+5)(2x^2-28x)-(4x-28)(x^3+5))/((2x^2-28x)^2)

undefined at (2x²-28x)², 2x² -28x=0

2x(x-14) = 0

x = 2,14

0 at (3x²+5)(2x²-28x)-(4x-28)(x³+5)

x = 0.761

User JohnSpeeks
by
8.3k points

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