What is the extreme f(x)=x^3-6x^2-15x+4 ?

The extreme f(x)=x^3-6x^2-15x+4 is Maximum(-1,12),Minimum(5,-96)

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extreme f(x)=x^3-6x^2-15x+4

extreme points

f (x) = x^{3} - 6 x^{2} - 15 x + 4

Maximum (- 1, 12), Minimum (5, - 96)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x - 15

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 1

into

x^{3} - 6 x^{2} - 15 x + 4 : 12

Maximum (- 1, 12)

Plug

x = 5

into

x^{3} - 6 x^{2} - 15 x + 4 : - 96

Minimum (5, - 96)

Maximum (- 1, 12), Minimum (5, - 96)

y = 2 - x

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

f (x) = x e^{2 x}

f (x) = x - 5

f (x) = - \frac{19}{( t + 6 ) ^{2}}

What is the extreme f(x)=x^3-6x^2-15x+4 ?
The extreme f(x)=x^3-6x^2-15x+4 is Maximum(-1,12),Minimum(5,-96)