How to Find the number of tangents to the curve y=$f\left(x\right)$ parallel to line $x+y=0$

Question

Question For $x$$>0,$ let

$f\left(x\right)=\int_{1}^{x}\left(\sqrt{\log t}-\frac{1}{2}\log\sqrt{t}\right)dt$

The number of tangents to the curve y=$f\left(x\right)$parallel to line x+y=0 is _________________

MY approach $x+y=0\Longrightarrow\frac{dy}{dx}=-1$

$f\left(x\right)=\int_{1}^{x}\left(\sqrt{\log t}-\frac{1}{2}\log\sqrt{t}\right)dt$$\Longrightarrow f'\left(x\right)=$$\sqrt{\log x}-\frac{1}{2}\log\sqrt{x}$

I know that parallel lines have same slope,but it would not give me number of tangents.

Answer 1

Hint: Finding the number of tangents of $f(x)$ parallel to line $y=-x$ is equivalent to finding the number of roots of the equation $f'(x)=-1$.

Answer 2

In each point where you will have $f'(x)=-1$ you will have such a tangent : \begin{align*} &-1=\sqrt{logx}-\frac{1}{2}log\sqrt{x} \end{align*} Since $f'$ seems to be a positive function (after a quick study of it), the answer must be $0$...