In which interval the function #f(x)=sqrt(x^2+8)-x# is a decreasing function?

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Answer 1 · 2017-04-04T05:23:13

#f(x)=sqrt(x^2+8)-x# is a decreasing function for all values of #x#

Intervals of increasing of a function are those where #f'(x)>0# and intervals of decreasing for a function are those where #f'(x)<0#.

Here #f(x)=sqrt(x^2+8)-x#

and #f'(x)=(df)/(dx)=1/(2sqrt(x^2+8))xx2x-1#

#=x/sqrt(x^2+8)-1#

#=(x-sqrt(x^2+8))/sqrt(x^2+8)#

It is apparent that as #x^2+8# is always positive and #sqrt(x^2+8)>x#,

#x-sqrt(x^2+8)# is negative for all values of #x#

i.e. for all #x# #f'(x)<0#.

Hence, #f(x)=sqrt(x^2+8)-x# is a decreasing function for all values of #x#

graph{sqrt(x^2+8)-x [-8.125, 11.875, -3.28, 6.72]}