Find Minimum value of $\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}$ Find Minimum value of $$f(x)=\sqrt{58-42x}+\sqrt{149-140\sqrt{1-x^2}}$$
My try: the domain of the function is $x \in [-1 \,\,\,1]$
Differentiating and equating it to zero we get
$$f'(x)=\frac{-21}{\sqrt{58-42x}}+\frac{70}{\sqrt{1-x^2}\sqrt{149-140\sqrt{1-x^2}}}=0$$
but its very tedious to find critical points here.
any other approach?
 A: hint
As $x\in[-1,1]$, you can put 
$$x=\cos(t)$$ with $$0\le t\le \pi.$$
the function becomes
$$F(t)=$$
$$\sqrt{58-42\cos(t)}+\sqrt{149-140\sin(t)}$$
$$\frac 17F'(t)=$$
$$\frac{3\sin(t)}{\sqrt{58-42\cos(t)}}-\frac{10\cos(t)}{\sqrt{149-140\sin(t)}}$$
$F'(t)=0$ gives
$$9\sin^2(t)(149-140\sin(t))=$$
$$100\cos^2(t)(58-42\cos(t))$$
A: Let $(x,y)$ be a point on the unit circle $x^2+y^2=1$. We have to minimize the function:
$$
\begin{aligned}
f(x,y)
&=
\sqrt{(7x-3)^2+(7y-0)^2} \ +\ \sqrt{(7x-0)^2+(7y-10)^2} 
\\
&=
\operatorname{Distance}(\ (7x,7y)\ ,\ (3,0)\ )\
+\
\operatorname{Distance}(\ (7x,7y)\ ,\ (0,10)\ )
\\
&\ge
\operatorname{Distance}(\ (3,0)\ ,\ (0,10)\ )
=\sqrt{3^2+10^2}\ ,
\end{aligned}
$$
with equality in the $\ge$ above only for the point of intersection of the segment with the above distance with the circle of radius $7$ centered in the origin.
A: If you are not interested in the analytical zero of $f'(x)$ you could try to find the solution using e.g. the Bisection method, or a Fixed-point iteration, e.g. Newton's method. However, if you are going to be solving it by hand, I admit that this doesn't really help with the tedious part – in fact, it is probably worse.
Edit: Also, as noted in a comment, your differentiation is not correct. Check your second term.
