What is the extreme f(x)=x^3-3/2 x^2-36x-5 ?

The extreme f(x)=x^3-3/2 x^2-36x-5 is Maximum(-3, 125/2),Minimum(4,-109)

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extreme f(x)=x^3-3/2 x^2-36x-5

extreme

Maximum (- 3, \frac{125}{2}), Minimum (4, - 109)

Solution steps

f^{'} (x) = 3 x^{2} - 3 x - 36

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 3

into

x^{3} - \frac{3}{2} x^{2} - 36 x - 5 : \frac{125}{2}

Maximum (- 3, \frac{125}{2})

Plug

x = 4

into

x^{3} - \frac{3}{2} x^{2} - 36 x - 5 : - 109

Minimum (4, - 109)

Maximum (- 3, \frac{125}{2}), Minimum (4, - 109)

What is the extreme f(x)=x^3-3/2 x^2-36x-5 ?
The extreme f(x)=x^3-3/2 x^2-36x-5 is Maximum(-3, 125/2),Minimum(4,-109)