It isn't familiar, right? - Part I

I don't need Calculus. From the conditional $-2\leq x\leq 5$ . We see $f(x)=\sqrt{-x^2+4x+21}-\sqrt{-x^2+3x+10}$ in the range $-2\leq x\leq 5$ $f(x')=\frac{-2x+4}{2\sqrt{-x^2+4x+21}}-\frac{-2x+3}{2\sqrt{-x^2+3x+10}}$ $f(x')=0\Leftrightarrow (-2x+4)\sqrt{-x^2+3x+10}=(-2x+3)\sqrt{-x^2+4x+21}$ $\Leftrightarrow x=\frac{1}{3}$ $f(-2)=3,f(\frac{1}{3})=\sqrt{2},f(5)=4$ So $f(x)_{min}=f(\frac{1}{3})=\sqrt{2}\approx 1.41$ Pls check back my solution and remove the report.

Since i knew nothing about derivative, here's my other solution

First, $x\in[-2;5]$

We rewrite the expression as $f(x)=\sqrt{(7-x)(x+3)}+\sqrt{(5-x)(x+2)}$ So $f(x)^2=(7-x)(x+3)+(5-x)(x+2)-2\sqrt{(7-x)(x+3)(5-x)(x+2)}$ $f(x)^2=(7-x)(x+2)+(5-x)(x+3)+7-x-5+x-2\sqrt{(7-x)(x+3)(5-x)(x+2)}$ $f(x)^2=\bigg[\sqrt{(7-x)(x+2)}-\sqrt{(5-x)(x+3)}\bigg]^2+2\geq 2$ $\Rightarrow f(x)_{min} = \sqrt{2}\approx 1.41\Leftrightarrow (7-x)(x+2)=(5-x)(x+3)\Leftrightarrow x=\frac{1}{3}$

Differentiating the function ans setting the resulting expression equal to 0 leads to the following quadratic equation : 51x^2 - 104x + 29 =0 with roots x 1 = 1.706 and x 2 = 1/3. Substituting the latter into f(1/3) yields sqrt(2). Ed Gray