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

The extreme f(x)=6-(x^3)/3+15x-x^2 is Minimum(-5,-157/3),Maximum(3,33)

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

extreme points

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

Minimum (- 5, - \frac{157}{3}), Maximum (3, 33)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 5

into

6 - \frac{x ^{3}}{3} + 15 x - x^{2} : - \frac{157}{3}

Minimum (- 5, - \frac{157}{3})

Plug

x = 3

into

6 - \frac{x ^{3}}{3} + 15 x - x^{2} : 33

Maximum (3, 33)

Minimum (- 5, - \frac{157}{3}), Maximum (3, 33)

f (x) = 3 x^{2} - 12 x + 7, 0 \leq x \leq 4

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

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

f (x) = 40 x^{2} + \frac{600}{x}

(y^{2} - x^{2}) (y - 2)

What is the extreme f(x)=6-(x^3)/3+15x-x^2 ?
The extreme f(x)=6-(x^3)/3+15x-x^2 is Minimum(-5,-157/3),Maximum(3,33)