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

The extreme f(x)=-sqrt(x^2+2x+5) is Maximum(-1,-2)

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

extreme points

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

Maximum (- 1, - 2)

Solution steps

f^{'} (x) = - \frac{x + 1}{x ^{2} + 2 x + 5}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 1

into

- x^{2} + 2 x + 5 : - 2

Maximum (- 1, - 2)

f (x) = \frac{1}{( x + 3 ) ^{\frac{1}{2}}}

f (x) = \frac{3}{x - 5}

\frac{2 x + 3}{x - 1}

\frac{5}{6} (x + 1)^{2} (x - 1) (x - 4)

f (x) = 3 x^{2} - x^{10}

What is the extreme f(x)=-sqrt(x^2+2x+5) ?
The extreme f(x)=-sqrt(x^2+2x+5) is Maximum(-1,-2)