What is the extreme f(x)= 1/(x^2-x-6) ?

The extreme f(x)= 1/(x^2-x-6) is Maximum(1/2 ,-4/25)

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extreme f(x)= 1/(x^2-x-6)

extreme points

f (x) = \frac{1}{x ^{2} - x - 6}

Maximum (\frac{1}{2}, - \frac{4}{25})

Solution steps

f^{'} (x) = - \frac{2 x - 1}{( x ^{2} - x - 6 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = \frac{1}{2}

into

\frac{1}{x ^{2} - x - 6} : - \frac{4}{25}

Maximum (\frac{1}{2}, - \frac{4}{25})

f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x - 5

8 x y - 2 x^{4} - 2 y^{4}

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 7

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

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

What is the extreme f(x)= 1/(x^2-x-6) ?
The extreme f(x)= 1/(x^2-x-6) is Maximum(1/2 ,-4/25)