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Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

 May 3, 2018
edited by Guest  May 3, 2018
edited by Guest  May 3, 2018

Best Answer 

 #4
avatar+22581 
+1

Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

 

 

\(\begin{array}{|rcll|} \hline f(x) &=& x + 2 \\ f(-3) &=& -3 + 2 \\ &=& -1 \\\\ g(x) &=& \frac{1}{f(x)} \\ g(f(-3)) &=& g(-1) \\ &=& \frac{1}{f(-1)} \quad & | \quad f(-1) = -1 + 2 = 1 \\ &=& \frac{1}{1} \\ &=& 1 \\\\ \mathbf{g(f(-3))} &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

 

laugh

 May 3, 2018
 #1
avatar
0

Answer is not -1 apparently...

 May 3, 2018
 #2
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0

Hint: g(x) is also, 1/x+2. 

 May 3, 2018
 #3
avatar+10413 
+1

Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

 

laugh

 May 3, 2018
 #4
avatar+22581 
+1
Best Answer

Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

 

 

\(\begin{array}{|rcll|} \hline f(x) &=& x + 2 \\ f(-3) &=& -3 + 2 \\ &=& -1 \\\\ g(x) &=& \frac{1}{f(x)} \\ g(f(-3)) &=& g(-1) \\ &=& \frac{1}{f(-1)} \quad & | \quad f(-1) = -1 + 2 = 1 \\ &=& \frac{1}{1} \\ &=& 1 \\\\ \mathbf{g(f(-3))} &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

 

laugh

heureka May 3, 2018

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