What is the extreme f(x)=-sqrt(x^2+8x+41) ?

The extreme f(x)=-sqrt(x^2+8x+41) is Maximum(-4,-5)

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extreme f(x)=-sqrt(x^2+8x+41)

extreme points

f (x) = - x^{2} + 8 x + 41

Maximum (- 4, - 5)

Solution steps

f^{'} (x) = - \frac{x + 4}{x ^{2} + 8 x + 41}

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < \infty

Plug

x = - 4

into

- x^{2} + 8 x + 41 : - 5

Maximum (- 4, - 5)

What is the extreme f(x)=-sqrt(x^2+8x+41) ?
The extreme f(x)=-sqrt(x^2+8x+41) is Maximum(-4,-5)