Find the extreme values of the function and where they occur? 1/x^2+1 i solve it as 0.5 is minimum at x= -1 but it is false!

Question

Find the extreme values of the function and where they occur? 1/x^2+1 i solve it as 0.5 is minimum at x= -1 but it is false!

1 Answer

Impact of this question

1960 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T15:58:19

$f [x] = 1$ $f[x] = 1$ when $x = 0$ $x=0$

Explanation:

$f [x]$ $f[x]$ = $\frac{1}{x^{2} + 1}$ $1/[x^2+1]$ ......Differentiating using the quotient rule.

$\frac{d}{d x} [\frac{u}{v}]$ $d/dx[u/v]$ = $\frac{\frac{v d u}{d x} - \frac{u d v}{d x}}{v^{2}}$ $[[vdu]/[dx]-[udv]/[dx]]/[v^2]$ = $\frac{[x^{2} + 1] [0] - [2 x]}{{[x^{2} + 1]}^{2}}$ $[[x^2+1][0] - [2x]]/[x^2+1]^2$ = - $\frac{2 x}{{[x^{2} + 1]}^{2}}$ $[2x]/[x^2+1]^2$ [ where $u$ $u$ and $v$ $v$ are both functions of $x$ $x$ , in this case $v = x^{2} + 1$ $v= x^2+1$ and $u = 1$ $u = 1$ ]

Therefore $- \frac{2 x}{{[x^{2} + 1]}^{2}}$ $- [2x]/[x^2+1]^2$ = $0$ $0$ for max/min. i.e, $x = 0$ $x=0$

When $x = 0$ $x=0$ , $f [x]$ $f[x]$ = $\frac{1}{{[0^{2} + 1]}^{2}}$ $1/[ 0^2+1]^2$ = $1$ $1$ .

Science

Math

Humanities

... and beyond

Find the extreme values of the function and where they occur? 1/x^2+1 i solve it as 0.5 is minimum at x= -1 but it is false!

1 Answer

Explanation:

Related questions

Impact of this question