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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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 Jun 15, 2019
 #1
avatar+999 
0

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

 Jun 15, 2019
 #2
avatar+8842 
+3

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)

 

\(g(x)\ =\ \dfrac{1}{f^{-1}(x)}+9\\~\\ g(x)\ =\ \dfrac{1}{2-\frac3x}+9\\~\\ g(3)\ =\ \dfrac{1}{2-\frac33}+9\\~\\ g(3)\ =\ \dfrac{1}{2-1}+9\\~\\ g(3)\ =\ 1+9\\~\\ g(3)\ =\ 10\)_

 Jun 15, 2019
 #3
avatar+106977 
+1
 Jun 16, 2019

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