What is the extreme f(x)=(-(1/2)x^4+2)/(3x^3-4) ?

The extreme f(x)=(-(1/2)x^4+2)/(3x^3-4) is Saddle(0,-1/2)

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extreme f(x)=(-(1/2)x^4+2)/(3x^3-4)

extreme

Saddle (0, - \frac{1}{2})

Solution steps

Find the critical points:

x = 0

Domain of

\frac{- \frac{1}{2} x ^{4} + 2}{3 x ^{3} - 4} : x < 3 \frac{4}{3} or x > 3 \frac{4}{3}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3 \frac{4}{3},

Decreasing

: 3 \frac{4}{3} < x < \infty

Plug the extreme point

x = 0

into

\frac{- \frac{1}{2} x ^{4} + 2}{3 x ^{3} - 4} \Rightarrow y = - \frac{1}{2}

Saddle (0, - \frac{1}{2})

What is the extreme f(x)=(-(1/2)x^4+2)/(3x^3-4) ?
The extreme f(x)=(-(1/2)x^4+2)/(3x^3-4) is Saddle(0,-1/2)