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

The extreme f(x)=(2x^2+1)/(x^2-4) is Maximum(0,-1/4)

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

extreme points

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

Maximum (0, - \frac{1}{4})

Solution steps

f^{'} (x) = - \frac{18 x}{( x ^{2} - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2} + 1}{x ^{2} - 4} : - \frac{1}{4}

Maximum (0, - \frac{1}{4})

f (t) = t \cdot e^{- ∣ x ∣}

f (x) = \frac{x ^{4}}{4} + x^{3} - 14 x^{2} - 60 x

f (x) = 4 x + z

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 70

f (x) = \frac{3}{4} x + 2, 0 \leq x \leq 4

What is the extreme f(x)=(2x^2+1)/(x^2-4) ?
The extreme f(x)=(2x^2+1)/(x^2-4) is Maximum(0,-1/4)