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

The extreme f(x)=(x+5)(x-3)^2 is Maximum(-7/3 , 2048/27),Minimum(3,0)

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

extreme points

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

Maximum (- \frac{7}{3}, \frac{2048}{27}), Minimum (3, 0)

Solution steps

f^{'} (x) = (x - 3) (3 x + 7)

Find intervals:

Increasing

: - \infty < x < - \frac{7}{3},

Decreasing

: - \frac{7}{3} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - \frac{7}{3}

into

(x + 5) (x - 3)^{2} : \frac{2048}{27}

Maximum (- \frac{7}{3}, \frac{2048}{27})

Plug

x = 3

into

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

Minimum (3, 0)

Maximum (- \frac{7}{3}, \frac{2048}{27}), Minimum (3, 0)

f (x) = x + e^{- 4 x}

f (x) = - 2 x^{3} + 9 x^{2} - 12 x

- \frac{5 x}{x ^{2} + 4}

f (x) = x^{2} + 6 x

f (x) = x^{2} + 5 x

What is the extreme f(x)=(x+5)(x-3)^2 ?
The extreme f(x)=(x+5)(x-3)^2 is Maximum(-7/3 , 2048/27),Minimum(3,0)