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

The extreme (7x(1+12x^2)^{-2}),-7<= x<= 12 is

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

extreme points

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

Maximum (- 7, - \frac{49}{346921}), Minimum (- \frac{1}{6}, - \frac{21}{32}), Maximum (\frac{1}{6}, \frac{21}{32}), Minimum (12, \frac{12}{427063})

Solution steps

Find the critical points:

x = - 7, x = - \frac{1}{6}, x = \frac{1}{6}, x = 12

Domain of

7 x (1 + 12 x^{2})^{- 2}, - 7 \leq x \leq 12 : - 7 \leq x \leq 12

Find intervals:

Decreasing

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

Increasing

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

Decreasing

: \frac{1}{6} < x < 12

Plug the extreme point

x = - 7

into

7 x (1 + 12 x^{2})^{- 2} \Rightarrow y = - \frac{49}{346921}

Maximum (- 7, - \frac{49}{346921})

Plug the extreme point

x = - \frac{1}{6}

into

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

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

Plug the extreme point

x = \frac{1}{6}

into

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

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

Plug the extreme point

x = 12

into

7 x (1 + 12 x^{2})^{- 2} \Rightarrow y = \frac{12}{427063}

Minimum (12, \frac{12}{427063})

Maximum (- 7, - \frac{49}{346921}), Minimum (- \frac{1}{6}, - \frac{21}{32}), Maximum (\frac{1}{6}, \frac{21}{32}), Minimum (12, \frac{12}{427063})

f (x) = \frac{x - 7}{x ^{2} - 25}, 0 \leq x < 5

f (x) = x 600 - x

f (x, y) = x (y + 1) + (x + y + 1)

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

f (x) = \frac{25}{2} x^{2} - ln (x)

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