What is the extreme f(x)=2x^3-39x^2+240x+8 ?

The extreme f(x)=2x^3-39x^2+240x+8 is Maximum(5,483),Minimum(8,456)

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extreme f(x)=2x^3-39x^2+240x+8

extreme points

f (x) = 2 x^{3} - 39 x^{2} + 240 x + 8

Maximum (5, 483), Minimum (8, 456)

Solution steps

f^{'} (x) = 6 x^{2} - 78 x + 240

Find intervals:

Increasing

: - \infty < x < 5,

Decreasing

: 5 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 5

into

2 x^{3} - 39 x^{2} + 240 x + 8 : 483

Maximum (5, 483)

Plug

x = 8

into

2 x^{3} - 39 x^{2} + 240 x + 8 : 456

Minimum (8, 456)

Maximum (5, 483), Minimum (8, 456)

f (x) = \frac{9 x ^{2} + x}{x ^{2} + 4}

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

f (x) = 3^{x} - 1

f (x) = \frac{( - x ^{2} + 8 )}{( 2 x ^{2} - 3 )}

f (x) = \frac{x}{x ^{2} + 16}

What is the extreme f(x)=2x^3-39x^2+240x+8 ?
The extreme f(x)=2x^3-39x^2+240x+8 is Maximum(5,483),Minimum(8,456)