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

The extreme f(x)=(x-5)^2(x+5) is Maximum(-5/3 , 4000/27),Minimum(5,0)

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

extreme points

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

Maximum (- \frac{5}{3}, \frac{4000}{27}), Minimum (5, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{5}{3},

Decreasing

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

Increasing

: 5 < x < \infty

Plug

x = - \frac{5}{3}

into

(x - 5)^{2} (x + 5) : \frac{4000}{27}

Maximum (- \frac{5}{3}, \frac{4000}{27})

Plug

x = 5

into

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

Minimum (5, 0)

Maximum (- \frac{5}{3}, \frac{4000}{27}), Minimum (5, 0)

f (x, y) = \frac{x ^{2} + y ^{2}}{x ^{2} y}

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

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

f (x) = x - sin (- x), 0 \leq x \leq 2 π

f (x) = \frac{( x ^{4} + 1 )}{x ^{2}}

What is the extreme f(x)=(x-5)^2(x+5) ?
The extreme f(x)=(x-5)^2(x+5) is Maximum(-5/3 , 4000/27),Minimum(5,0)