What is the extreme f(x)=(27)/(x^2-4x-5) ?

The extreme f(x)=(27)/(x^2-4x-5) is Maximum(2,-3)

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extreme f(x)=(27)/(x^2-4x-5)

extreme points

f (x) = \frac{27}{x ^{2} - 4 x - 5}

Maximum (2, - 3)

Solution steps

f^{'} (x) = - \frac{27 ( 2 x - 4 )}{( x ^{2} - 4 x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 2,

Decreasing

: 2 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = 2

into

\frac{27}{x ^{2} - 4 x - 5} : - 3

Maximum (2, - 3)

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x = \frac{4}{3}

y = x^{5} - x^{4}

f (x, y) = 49 - (x - 9)^{2} - y^{2}

f (x, y) = x^{2} y^{2} + 2 x y

f (x) = 4 x^{2} - 5 x - 2

What is the extreme f(x)=(27)/(x^2-4x-5) ?
The extreme f(x)=(27)/(x^2-4x-5) is Maximum(2,-3)