1. If $f(x) = x^2 - 1$ and $g(x) = x + 7,$ evaluate $f(g(7)) + g(f(3)).$
2. Suppose $f(x)=\frac{3}{2-x}$. If $g(x)=\frac{1}{f^{-1}(x)}+9$, find $g(3)$.
3. For what value of does the equation represent a circle of radius 6?
4. The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers. Find $a-b+c$.
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+1253 f(g(7))+g(f(3)). Seems complex at first.
However, we can evaluate g(7), which is 7+7 = 14, and now we can find f(14)
Also, we can evaluate f(3), which is 3^2-1=8, and now we can find g(8).
f(14)=14^2-1=195 and g(8)=8+7=15.
Now we have 195+15 which equals 210.
You are very welcome!
:P
+9466 2. Suppose \(f(x)=\frac{3}{2-x}\) . If \(g(x)=\frac{1}{f^{-1}(x)}+9\) , find \(g(3)\) .
Let's find f-1(x) by setting y = f(x) and solving for x .
\(y\ =\ \frac{3}{2-x}\\~\\ y(2-x)\ =\ 3\qquad\text{and}\qquad x\neq2\\~\\ 2-x\ =\ \frac3y\\~\\ -x\ =\ -2+\frac3y\\~\\ x\ =\ 2-\frac3y\)
So we know...
\(f^{-1}(x)\ =\ 2-\frac3x\)
And we can substitute this in for f-1(x) in the equation for g(x)
\(g(x)\ =\ \dfrac{1}{f^{-1}(x)}+9\\~\\ g(x)\ =\ \dfrac{1}{2-\frac3x}+9\\~\\ g(3)\ =\ \dfrac{1}{2-\frac33}+9\\~\\ g(3)\ =\ \dfrac{1}{2-1}+9\\~\\ g(3)\ =\ 1+9\\~\\ g(3)\ =\ 10\)_
