I was trying to find the range of the function
$$f(x) = \frac{3}{4-5\sin(x)} $$
The first thought that came to mind was:
Let's put the minimum and maximum value of $\sin(x)$ as $\{1,-1\}$ accordingly to find maximum and minimum of the function.
However, then I got range $[-3,\frac{1}{3}]$, which was totally wrong when I used a graphing calculator.
The true answer is $(-\infty,-3] ∪ [\frac{1}{3}, \infty) $
Now, I want explaination what I did wrong here.
Best Answer
Let the denominator be $y$. Then, as you observed $-1\leq y\leq 9$. But you have to be careful when find the range for $\frac 1 y$. At $y=0$, $\frac 1 y$ is not defined. You have to consider $0<y\leq 9$ and $-1\leq y<0$ separately. For $0<y\leq 9$ we get $\frac 1 y \in [\frac 1 9, \infty)$ and for $-1\leq y<0$ we get $\frac 1 y \in (-\infty, -1]$. Now multiply by $3$.