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

1 Answer
Apr 4, 2017

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

Explanation:

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]}