Step by step solution to "f(x)=xx2+1 Find the absolute extreme values on [-7,7]"

pancha3

Answered question

2021-08-13

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

Find the absolute extreme values on [-7,7]

Answer & Explanation

komunidadO

Skilled2021-08-14Added 86 answers

First direction:
$f (x) = \frac{x}{x^{2} + 1}$ on [-7,7]
$f^{'} (x) = \frac{d}{d x} (\frac{x}{x^{2} + 1})$
$= \frac{(x^{2} + 1) - x (2 x)}{{(x^{2} + 1)}^{2}}$
$= \frac{1 - x^{2}}{{(x^{2} + 1)}^{2}}$
Equate above first derivative to zero to obtain the extreme points:
$f^{'} (x) = 0$
$\frac{1 - x^{2}}{{(x^{2} + 1)}^{2}} = 0$
$1 - x^{2} = 0$
$x^{2} = 1$
$x = \pm 1$
Compute the second derivative:
$f^{'} (x) = \frac{1 - x^{2}}{{(x^{2} + 1)}^{2}}$
$f^{″} (x) = \frac{d}{d x} (\frac{1 - x^{2}}{{(x^{2} + 1)}^{2}})$
$= \frac{{(x^{2} + 1)}^{2} (- 2 x) - (1 - x^{2}) (4 x (x^{2} + 1))}{{(x^{2} + 1)}^{4}}$
$f^{″} (1) = \frac{1 (- 2) - 0}{1} = - 2 < 0$ [max at 1]
$f^{″} (- 1) = \frac{1 (2) - 0}{1} = 2 < 0$ [min at -1]
Thus,
Max value is $f (1) = \underset{o f 1}{=} (\frac{x}{x^{2} + 1}) = \frac{1}{2}$
Min value is $f (- 1) = \underset{o f - 1}{=} (\frac{x}{x^{2} + 1}) = - \frac{1}{2}$

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