What is the extreme (x^2-x)/(8x^2+1) ?

The extreme (x^2-x)/(8x^2+1) is Maximum(-1/2 , 1/4),Minimum(1/4 ,-1/8)

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extreme (x^2-x)/(8x^2+1)

extreme points

\frac{x ^{2} - x}{8 x ^{2} + 1}

Maximum (- \frac{1}{2}, \frac{1}{4}), Minimum (\frac{1}{4}, - \frac{1}{8})

Solution steps

f^{'} (x) = \frac{8 x ^{2} + 2 x - 1}{( 8 x ^{2} + 1 ) ^{2}}

Find intervals:

Increasing

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

Decreasing

: - \frac{1}{2} < x < \frac{1}{4},

Increasing

: \frac{1}{4} < x < \infty

Plug

x = - \frac{1}{2}

into

\frac{x ^{2} - x}{8 x ^{2} + 1} : \frac{1}{4}

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

Plug

x = \frac{1}{4}

into

\frac{x ^{2} - x}{8 x ^{2} + 1} : - \frac{1}{8}

Minimum (\frac{1}{4}, - \frac{1}{8})

Maximum (- \frac{1}{2}, \frac{1}{4}), Minimum (\frac{1}{4}, - \frac{1}{8})

(5 - x)^{3}

f (x) = y^{x}

f (x) = - x^{2} + 3 x - 5

f (x) = 2 x^{3} + 2 y^{2} - 6 x - 8 y + 14

f (x) = 2 x^{2} - 8 [0.6]

What is the extreme (x^2-x)/(8x^2+1) ?
The extreme (x^2-x)/(8x^2+1) is Maximum(-1/2 , 1/4),Minimum(1/4 ,-1/8)