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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

#4
+20866
+1

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

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

May 3, 2018

#1
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May 3, 2018
#2
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Hint: g(x) is also, 1/x+2.

May 3, 2018
#3
+9807
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Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?

May 3, 2018
#4
+20866
+1

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

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

heureka May 3, 2018