Question

What are the critical points of `f(x) = sqrt(e^(sqrtx)-sqrtx)`?

Answer 1

`x=0`

Explanation:

`f(x)=(e^(x^(1/2))-x^(1/2))^(1/2)`

Through the chain rule:

`f'(x)=1/2(e^(x^(1/2))-x^(1/2))^(-1/2)d/dx(e^(x^(1/2))-x^(1/2))`

Then:

`f'(x)=1/2(e^(x^(1/2))-x^(1/2))^(-1/2)(e^(x^(1/2))(1/2x^(-1/2))-1/2x^(-1/2))`

Factoring from the final parentheses:

`f'(x)=1/2(e^(x^(1/2))-x^(1/2))^(-1/2)(1/2x^(-1/2))(e^(x^(1/2))-1)`

Rewriting:

`f'(x)=1/(2(e^(x^(1/2))-x^(1/2))^(1/2)(2x^(1/2)))(e^(x^(1/2))-1)`

`f'(x)=(e^sqrtx-1)/(4sqrtxsqrt(e^sqrtx-sqrtx))`

If we want to find critical point, we need to find when `f'(x)=0` or when `f'` is undefined but `f` is defined.

Setting `f'(x)=0` gives `e^sqrtx-1=0=>e^sqrtx=1=>x=0`.

This is also when `f'` is undefined, since `sqrtx` is in the denominator. We also see that `sqrt(e^sqrtx-sqrtx)` is never undefined, as `e^sqrtx>sqrtx` for all `x>=0`.

Thus the only critical point is `x=0`. It's also one of the function's endpoints.