Find-the-maximum-value-of-the-function-f-defined-by-f-x-x-1-x-2-x-R-

Question Number 83850 by Rio Michael last updated on 06/Mar/20

Find the maximum value of the function f, defined by

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

Commented by mathmax by abdo last updated on 06/Mar/20

{l i m}_{x \to \infty} f (x) = 0 a n d f^{^{'}} (x) = \frac{1 + x^{2} - x (2 x)}{{(1 + x^{2})}^{2}} = \frac{1 - x^{2}}{{(1 + x^{2})}^{2}}

w e h a v e f^{^{'}} (x) ⩾ 0 \Leftrightarrow - 1 ⩽ x ⩽ 1

x - \infty - 1 0 1 + \infty

f^{^{'}} - + + -

f 0 d e c r - \frac{1}{2} i n c 0 i n c \frac{1}{2} d e c 0

\Rightarrow {m a x}_{x \in R} f (x) = \frac{1}{2}

Answered by john santu last updated on 06/Mar/20

(x^{2} + 1) y - x = 0

{yx}^{2} - x + y = 0

\Rightarrow Δ = 1 - 4 . y^{2} ⩾ 0

(2 y - 1) (2 y + 1) ⩽ 0

- \frac{1}{2} ⩽ y ⩽ \frac{1}{2}

{\begin{cases} \max = \frac{1}{2} \\ \min = - \frac{1}{2} \end{cases}

Answered by mr W last updated on 07/Mar/20

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

⩽ \frac{1}{2} \to m a x .

⩾ \frac{1}{- 2} = - \frac{1}{2} \to m i n .

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