For the function f(x)=1/(x^2-2x-15), determine the range. I got yER|y>0 or y<=-0.0625. However, the book's answer is different from mine. Is mine correct?
Oh, of course! My mistake entirely. See the graph, it will help. I basically ignored the sections either side of the asymptotes. Consider, for example, $\displaystyle x=6$ or $\displaystyle x=-4$
There is a maximum point at $\displaystyle x=\frac{-1}{16}$, but the only impossible value for the range is when $\displaystyle y=0$ because you can never obtain such a value from the function.
Edit: Correction.
No, Look at my example of $\displaystyle x=-4$
$\displaystyle h(-4)=\frac{1}{(-4)^2-2(-4)-15}$
$\displaystyle =\frac{1}{16+8-15}$
$\displaystyle =\frac{1}{9}$
You have found a maximum point in the graph. If you use the link I provided above, and find the point, you can clearly see that it is a maximum point. What I mean by "maximum point" is not "this is the highest possible value the function can obtain", rather "at this point, the function stops increasing and starts decreasing." This stops, however, at the asymptote at $\displaystyle x=5$ At this point, the function exceeds the maximum point previously calculated; it has nothing to do therefore with the range of the function.
I'm waffling superfluously and confusingly; the core point is this:
To calculate the possible range of this function, you need to consider the values $\displaystyle h(x)$ cannot possibly take. In this case, there is just one such value: where $\displaystyle h(x)=0$. The maximum value is not necessary to find the range in this scenario.
Edit:correction.
Actually that's true and stumped me. You were right from the start, I think I completely misunderstood what you were doing.
It's clear that $\displaystyle h(x)>0$ or $\displaystyle h(x)\leq{\frac{-1}{16}}$
I'm wondering if this hole in the graph hasn't been considered because of a mistake. Otherwise, I have no clue. Someone else will hopefully be able to provide a more resolute answer.