1

$f: \mathbb{R} \to (-\infty, 4]$ $f(x) =-\frac {sin(\frac {11x\pi}{6})}{5}+2$

I don't quite understand how to prove that this is a surjective function. I know that all values of x are mapped to at least one value of y in the given co-domain. From what I understand I first have to solve for x, and then insert that value into the function to get y. But I just end up with a messy arcsine function. When I isolate x i get:

$x = \frac {arcsine(10-5y)}{11\pi}$

$f(\frac {arcsine(10-5y)}{11\pi}) = -\frac {sin(\frac {11\cdot \frac {arcsine(10-5y)}{11\pi}\pi}{6})}{5}+2$

(This function is terribly confusing).

Bullerskydd
  • 39

1 Answers1

1

Given $y \in (-\infty,4]$ can you find $x \in \mathbb{R}$ such that $y = f(x)$? No, so this function isn't surjective onto $(-\infty,4]$.

Because $-1 \leq \sin(\text{whatever}) \leq 1$, then $\frac{9}{5} \leq -\frac{\sin(\text{whatever})}{5}+2 \leq \frac{11}{5}$ and you can't find an $x$ such that, for example, $f(x)=1$.

dcolazin
  • 2,242