What is the extreme f(x)= 3/(4x^4-8x^2) ?

The extreme f(x)= 3/(4x^4-8x^2) is Maximum(-1,-3/4),Maximum(1,-3/4)

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extreme f(x)= 3/(4x^4-8x^2)

extreme

Maximum (- 1, - \frac{3}{4}), Maximum (1, - \frac{3}{4})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 1

into

\frac{3}{4 x ^{4} - 8 x ^{2}} : - \frac{3}{4}

Maximum (- 1, - \frac{3}{4})

Plug

x = 1

into

\frac{3}{4 x ^{4} - 8 x ^{2}} : - \frac{3}{4}

Maximum (1, - \frac{3}{4})

Maximum (- 1, - \frac{3}{4}), Maximum (1, - \frac{3}{4})

What is the extreme f(x)= 3/(4x^4-8x^2) ?
The extreme f(x)= 3/(4x^4-8x^2) is Maximum(-1,-3/4),Maximum(1,-3/4)