What is the extreme f(x)=sqrt(26x^2+80x+64) ?

The extreme f(x)=sqrt(26x^2+80x+64) is Minimum(-20/13 ,(4sqrt(2))/(sqrt(13)))

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extreme f(x)=sqrt(26x^2+80x+64)

extreme points

f (x) = 26 x^{2} + 80 x + 64

Minimum (- \frac{20}{13}, \frac{4 2}{13})

Solution steps

f^{'} (x) = \frac{2 ( 13 x + 20 )}{26 x ^{2} + 80 x + 64}

Find intervals:

Decreasing

: - \infty < x < - \frac{20}{13},

Increasing

: - \frac{20}{13} < x < \infty

Plug

x = - \frac{20}{13}

into

26 x^{2} + 80 x + 64 : \frac{4 2}{13}

Minimum (- \frac{20}{13}, \frac{4 2}{13})

f (x, y) = x^{3} - y^{2} - 12 x + 4 y + 2

2 x - \frac{360}{x ^{2}}

f (x) = (3 x - 3)^{2}, - \infty < x \leq 2

f (x, y, z) = x ln (y)

f (x, y) = 200 - x - y

What is the extreme f(x)=sqrt(26x^2+80x+64) ?
The extreme f(x)=sqrt(26x^2+80x+64) is Minimum(-20/13 ,(4sqrt(2))/(sqrt(13)))