What is the extreme f(x)=x^3-6x^2-15x+1 ?

The extreme f(x)=x^3-6x^2-15x+1 is Maximum(-1,9),Minimum(5,-99)

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extreme f(x)=x^3-6x^2-15x+1

extreme points

f (x) = x^{3} - 6 x^{2} - 15 x + 1

Maximum (- 1, 9), Minimum (5, - 99)

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 + 1 : 9

Maximum (- 1, 9)

Plug

x = 5

into

x^{3} - 6 x^{2} - 15 x + 1 : - 99

Minimum (5, - 99)

Maximum (- 1, 9), Minimum (5, - 99)

\frac{4 x ^{2}}{x ^{2} - 4}

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

f (x) = 2 x^{3} - 21 x^{2} + 60 x, 0 \leq x \leq 6

f (x) = e^{x^{2} - 6 x - 1}

What is the extreme f(x)=x^3-6x^2-15x+1 ?
The extreme f(x)=x^3-6x^2-15x+1 is Maximum(-1,9),Minimum(5,-99)