Why is the range of $D(x) = \frac{(5x+8)^2}{(x-15)^2}$ all non-negative numbers? $$D(x) = \frac{(5x+8)^2}{(x-15)^2}$$
The domain = all real numbers excluding $15$, because we cannot divide by $0$.
But, why is the range of this all real numbers greater than or equal to 0?
I've tried setting $D(x) = y$ and solving for $x$, but it's quite a tricky thing to solve for $x$. Any easier methods that you guys can suggest?
It is clear that for all $x \in \mathbb{R} \setminus \lbrace 15 \rbrace$, you have $D(x) \geq 0$. So the range is included in $\mathbb{R}_+$.
Moreover, $$\lim_{x \rightarrow 15^-} D(x) = +\infty \quad \text{ and } \quad D \left( \frac{-8}{5}\right) = 0$$ By continuity on $(-\infty, 15)$, the intermediate value theorem tells you that $\mathbb{R}_+$ is included in the range.
So the range of $D$ is $\mathbb{R}_+$.
Write
$$
D(x) = \frac{(5x+8)^2}{(x-15)^2} = \left(\frac{5x+8}{x-15}\right)^2 = f(x)^2
$$
The image of $f$ is $\mathbb R \setminus \{5\}$ because it is easy to solve $f(x)=a$ when $a\ne 5$. Therefore, image of $D$ is $\mathbb [0,\infty)$ except possibly for $25$. Since the image $f$ contains $-5$, the image of $D$ contains $(-5)^2=25$ and so is actually $[0,\infty)$.