Question

for f(x) = 1/(x^2-3), find f(2+h)

Answer 1

we have to replace \(x=2+h\), so we get:
\[f\left( {2 + h} \right) = \frac{1}{{{{\left( {2 + h} \right)}^2} - 3}} = ...?\]

Answer 2

I'm pretty lost on how to do this, I have my equation set up just like that, but the )2+h) part throws me off so much

Answer 3

(2+h)**

Answer 4

hint:
if we develop the square of binomial, we can write this:
\[\begin{gathered}
f\left( {2 + h} \right) = \frac{1}{{{{\left( {2 + h} \right)}^2} - 3}} = \hfill \\
\hfill \\
= \frac{1}{{4 + {h^2} - 4h - 3}} = ...? \hfill \\
\end{gathered} \]

Answer 5

(2+h)^2=4+h^2+4h

Answer 6

oops.. I have made a typo:
\[\begin{gathered}
f\left( {2 + h} \right) = \frac{1}{{{{\left( {2 + h} \right)}^2} - 3}} = \hfill \\
\hfill \\
= \frac{1}{{4 + {h^2} + 4h - 3}} = ...? \hfill \\
\end{gathered} \]

Answer 7

Should I have ended up with \[1\div h^2+4h-3\]?

Answer 8

I feel like I'm doing this very wrong

Answer 9

no, please you have to simplify similar terms, so we have to compute: 4-3=...?

Answer 10

4-3=1

Answer 11

correct! so, here is the final answer:
\[\begin{gathered}
f\left( {2 + h} \right) = \frac{1}{{{{\left( {2 + h} \right)}^2} - 3}} = \hfill \\
\hfill \\
= \frac{1}{{4 + {h^2} + 4h - 3}} = \hfill \\
\hfill \\
= \frac{1}{{{h^2} + 4h + 1}} \hfill \\
\end{gathered} \]

Answer 12

Thank you!!!

Answer 13

:)