What is the extreme f(x)=sqrt(10x^2+36x+36) ?

The extreme f(x)=sqrt(10x^2+36x+36) is Minimum(-9/5 ,(3sqrt(2))/(sqrt(5)))

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extreme f(x)=sqrt(10x^2+36x+36)

extreme points

f (x) = 10 x^{2} + 36 x + 36

Minimum (- \frac{9}{5}, \frac{3 2}{5})

Solution steps

f^{'} (x) = \frac{2 ( 5 x + 9 )}{10 x ^{2} + 36 x + 36}

Find intervals:

Decreasing

: - \infty < x < - \frac{9}{5},

Increasing

: - \frac{9}{5} < x < \infty

Plug

x = - \frac{9}{5}

into

10 x^{2} + 36 x + 36 : \frac{3 2}{5}

Minimum (- \frac{9}{5}, \frac{3 2}{5})

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

f (x) = 4 x^{3} + 48 x^{2} + 180 x

f (x) = \frac{x ^{2} - 7 x + 26}{x - 5}

f (x) = x^{3} - x^{2} - 8 x + 4

f (x) = x^{3} - x^{2} - 8 x + 8

What is the extreme f(x)=sqrt(10x^2+36x+36) ?
The extreme f(x)=sqrt(10x^2+36x+36) is Minimum(-9/5 ,(3sqrt(2))/(sqrt(5)))