What is the extreme f(x)= 1/(x^2-4x-21) ?

The extreme f(x)= 1/(x^2-4x-21) is Maximum(2,-1/25)

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extreme f(x)= 1/(x^2-4x-21)

extreme points

f (x) = \frac{1}{x ^{2} - 4 x - 21}

Maximum (2, - \frac{1}{25})

Solution steps

f^{'} (x) = - \frac{2 x - 4}{( x ^{2} - 4 x - 21 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 2,

Decreasing

: 2 < x < 7,

Decreasing

: 7 < x < \infty

Plug

x = 2

into

\frac{1}{x ^{2} - 4 x - 21} : - \frac{1}{25}

Maximum (2, - \frac{1}{25})

P (x, y) = 16 x^{2} - 9 y^{2}

f (x) = 1920 x - 10 x^{3}

f (x) = 100 e^{- 0.062 \cdot x}

f (x) = 7 sin^{2} (x) - 14 cos (x)

f (x) = - 3 x^{2} - 30 x - 21

What is the extreme f(x)= 1/(x^2-4x-21) ?
The extreme f(x)= 1/(x^2-4x-21) is Maximum(2,-1/25)