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)

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

extreme points

f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x - 5

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)

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 7

f (x) = \frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + 2

f (x, y) = x^{2} + 2 y^{2} - 2 x - 4 y + 1

f (x, y) = 2 x^{2} + y^{2} + 8 x - 6 y - 2 x y + 12

f (x) = x^{3} - 3 x^{2} + 24 x

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)