Question

Is `f(x)=e^xsqrt(x^2-x)` increasing or decreasing at `x=3`?

Answer 1

`f(x)=e^xsqrt(x^2-x)` is increasing at `x=3`

Explanation:

A function `f(x)` is increasing at `x=a` if `f'(a)>0` and is decreasing at `x=a` if `f'(a)<0`.

As `f(x)=e^xsqrt(x^2-x)`, using product rule

`f'(x)=e^xsqrt(x^2-x)+e^x xx1/(2sqrt(x^2-x))xx(2x-1)`

= `e^xsqrt(x^2-x)+(e^x(2x-1))/(2sqrt(x^2-x))`

and at `x=3`

`f'(x)=e^3sqrt(3^2-3)+(e^3(2xx3-1))/(2sqrt(3^2-3))`

= `e^3sqrt6+(5e^3)/(2sqrt6)=e^3sqrt6(1+5/12)=e^3sqrt6xx17/12`

which is clearly positive.

Hence, `f(x)=e^xsqrt(x^2-x)` is increasing at `x=3`.