What is the extreme f(x)=-x^3+3x^2+144x+1 ?

The extreme f(x)=-x^3+3x^2+144x+1 is Minimum(-6,-539),Maximum(8,833)

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extreme f(x)=-x^3+3x^2+144x+1

extreme points

f (x) = - x^{3} + 3 x^{2} + 144 x + 1

Minimum (- 6, - 539), Maximum (8, 833)

Solution steps

f^{'} (x) = - 3 x^{2} + 6 x + 144

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 8,

Decreasing

: 8 < x < \infty

Plug

x = - 6

into

- x^{3} + 3 x^{2} + 144 x + 1 : - 539

Minimum (- 6, - 539)

Plug

x = 8

into

- x^{3} + 3 x^{2} + 144 x + 1 : 833

Maximum (8, 833)

Minimum (- 6, - 539), Maximum (8, 833)

3 x - 3 y + 12 = 0, at 1, - 1

f (x) = x + 6

ln (x)

f (x) = x^{2} - 121

(50, - 6), (2, 6)

What is the extreme f(x)=-x^3+3x^2+144x+1 ?
The extreme f(x)=-x^3+3x^2+144x+1 is Minimum(-6,-539),Maximum(8,833)