What is the extreme f(x)=(3-x)(x-5)^2 ?

The extreme f(x)=(3-x)(x-5)^2 is Minimum(11/3 ,-32/27),Maximum(5,0)

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extreme f(x)=(3-x)(x-5)^2

extreme points

f (x) = (3 - x) (x - 5)^{2}

Minimum (\frac{11}{3}, - \frac{32}{27}), Maximum (5, 0)

Solution steps

f^{'} (x) = - (x - 5) (3 x - 11)

Find intervals:

Decreasing

: - \infty < x < \frac{11}{3},

Increasing

: \frac{11}{3} < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = \frac{11}{3}

into

(3 - x) (x - 5)^{2} : - \frac{32}{27}

Minimum (\frac{11}{3}, - \frac{32}{27})

Plug

x = 5

into

(3 - x) (x - 5)^{2} : 0

Maximum (5, 0)

Minimum (\frac{11}{3}, - \frac{32}{27}), Maximum (5, 0)

(X^{2} - 1) (X^{2} - 9)

f (x) = \frac{x}{8 x + 3}, 8 \leq x \leq 12

f (x) = (x^{2} + 4) (16 - x^{2})

f (x) = x^{4} + x^{3} - 3 x^{2} + 1

f (x) = 9

What is the extreme f(x)=(3-x)(x-5)^2 ?
The extreme f(x)=(3-x)(x-5)^2 is Minimum(11/3 ,-32/27),Maximum(5,0)