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

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})

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)))