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
Barry H.
Jun 25, 2018

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

Explanation:

#f[x]#=#1/[x^2+1]#......Differentiating using the quotient rule.

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

Therefore #- [2x]/[x^2+1]^2# = # 0# for max/min. i.e, #x=0#

When #x=0#, #f[x]# = #1/[ 0^2+1]^2# = #1#.

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