What is the extreme f(x)=sqrt(x^2+6x+34) ?

The extreme f(x)=sqrt(x^2+6x+34) is Minimum(-3,5)

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extreme f(x)=sqrt(x^2+6x+34)

extreme points

f (x) = x^{2} + 6 x + 34

Minimum (- 3, 5)

Solution steps

f^{'} (x) = \frac{x + 3}{x ^{2} + 6 x + 34}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < \infty

Plug

x = - 3

into

x^{2} + 6 x + 34 : 5

Minimum (- 3, 5)

f (x) = \frac{( x + 2 )}{x - 3}

\frac{e ^{x} + e ^{- x}}{2}

(0) (14.14)

f (x) = - 3 cos (\frac{1}{3} x - π) + 4

f (x) = \frac{x ^{2} - x - 12}{2 x - 8}

What is the extreme f(x)=sqrt(x^2+6x+34) ?
The extreme f(x)=sqrt(x^2+6x+34) is Minimum(-3,5)