Find the product of maximum and minimum value of $f(x)=\sqrt{2}\sin{x}-f(0)$

Question

$f(x)=\sqrt{2}\sin{x}-f(0)$

(A) $\frac{-3\sqrt{2}}{4}$

(B) $\frac{-3\sqrt{2}}{2}$

(C) $\frac{-3}{2}$

(D) $\frac{-3}{4}$

(E) $\frac{-7}{4}$

My attempt :

let x=0

$f(0)=\sqrt{2}\sin{0}-f(0)\\ f(0)=-f(0)\\ 2f(0)=0\\ f(0)=0$

then :

$f(x)=\sqrt{2}\sin{x}$

the answer may be $-\sqrt{2}$ but it's not in the option.

Answer 1

You've done the tough bit!

Now remember that $-1\leq \sin x\leq 1$, so what would be the bounds of $\sqrt2 \sin x$?

As it turns out, none of the options you are given are correct.