I can't use the mean value theorem to prove this. The problem that I am given is $$ f(x) = (\sqrt{17\pi} )x^2 $$
on the interval $=-10 \le x \le 4$
I know that I have to show that $\lvert f(x_2)-f(x_1) \rvert \le k \lvert x_2-x_1 \rvert$
I start the process by simply plugging the value of f(x) that I am given into the equation above. I get $\lvert f(x_2)-f(x_1) \rvert = \lvert x_2^2 \sqrt{17\pi}-x_1^2\sqrt{17\pi} \rvert$. I then multiplied it by $\frac{x_2^2 \sqrt{17\pi}+x_1^2\sqrt{17\pi}}{x_2^2 \sqrt{17\pi}+x_1^2\sqrt{17\pi}}$
After much simplification I have
$$ \frac{17\pi}{\sqrt{17\pi}} \lvert\frac{X_2^4-X_1^4}{X_2^2-X_1^2}\rvert $$
I was just wondering if we have to have $\lvert f(x_2)-f(x_1) \rvert \le k \lvert x_2-x_1 \rvert$ or can I have $\lvert f(x_2)-f(x_1) \rvert \le k \lvert x_2^4-x_1^4 \rvert$?
I will give for general case: consider $f(x)=cx^2$ over the interval $[a,b]$. Let $M=\max(|a|,|b|)$ then for $x_1,x_2\in[a,b]$, $$|cx_1^2-cx_2^2|\le |c||x_1+x_2||x_1-x_2| \le 2|c|M|x_1-x_2|$$ since $|x_1+x_2|\le |x_1|+|x_2| \le 2M$.