What is the extreme 7x(1+12x^2)^{-2} ?

The extreme 7x(1+12x^2)^{-2} is Minimum(-1/6 ,-21/32),Maximum(1/6 , 21/32)

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extreme 7x(1+12x^2)^{-2}

extreme points

7 x (1 + 12 x^{2})^{- 2}

Minimum (- \frac{1}{6}, - \frac{21}{32}), Maximum (\frac{1}{6}, \frac{21}{32})

Solution steps

f^{'} (x) = \frac{7 ( - 36 x ^{2} + 1 )}{( 12 x ^{2} + 1 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{6},

Increasing

: - \frac{1}{6} < x < \frac{1}{6},

Decreasing

: \frac{1}{6} < x < \infty

Plug

x = - \frac{1}{6}

into

7 x (1 + 12 x^{2})^{- 2} : - \frac{21}{32}

Minimum (- \frac{1}{6}, - \frac{21}{32})

Plug

x = \frac{1}{6}

into

7 x (1 + 12 x^{2})^{- 2} : \frac{21}{32}

Maximum (\frac{1}{6}, \frac{21}{32})

Minimum (- \frac{1}{6}, - \frac{21}{32}), Maximum (\frac{1}{6}, \frac{21}{32})

f (x) = 4 sin (x) + 3 cos (x), 0 \leq x \leq 2 π

f (x, y) = 2 x^{2} + 4 x y - x^{2} y - 4 x

5 x^{2} (x + 1)^{\frac{1}{2}}

f (x) = y e^{x} + x y^{2}

f (x) = x^{3} - 8 y^{3} + 4 x y - 8

What is the extreme 7x(1+12x^2)^{-2} ?
The extreme 7x(1+12x^2)^{-2} is Minimum(-1/6 ,-21/32),Maximum(1/6 , 21/32)