Creating and Evaluating Composite Functions Practice II

1. The graphs of the equations ‍$y=f(x)$ and ‍‍$y=g(x)$ are shown in the grid below.

   
   

   Which of the following best approximates the value of ‍$g(f(-2))$?

   Choose 1 answer:
   

   1. -3
   2. -2
   3. 0
   4. 3
      

2. The tables below show some inputs and outputs of functions ‍$g$ and ‍$h$.

   $x$	-6	-4	-2	-1	0	2
   $g(x)$	-21	3	11	9	3	-21

   
   

   $x$	-1	0	2	3	4	6
   $h(x)$	5	2	2	5	10	26

   
   

   Evaluate.

   $(h\circ g) (0)=$
   

3. The tables below show some inputs and outputs of functions ‍$f$ and ‍$h$.

   $x$	-6	-4	-2	0	2	4
   $f(x)$	-4	-3	-2	-1	0	1

   
   

   $x$	-8	-4	-2	0	1	2
   $h(x)$	80	2	2	10	17	26

   
   

   Evaluate.

   $h(f(-2))=$
   

4. The graphs of the equations ‍$y=f(x)$ and ‍‍$y=g(x)$ are shown in the grid below.

   
   

   Which of the following best approximates the value of ‍$(f\circ g) (3)$?

   Choose 1 answer:
   

   1. -6
   2. -2
   3. 0
   4. 6

Source: Khan Academy, https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:composite/x9e81a4f98389efdf:composing/e/evaluate-composite-functions-from-graphs-and-tables
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1. Strategy
   

   We take the output from the inner function, $f$, and input it into the outer function, $g$.
   

   To evaluate $g({f(-2)})$, let's first evaluate ‍${f(-2)}$. Then we'll input that result into \(\) to find the output value.

   Evaluating ${f(-2)}$

   The best approximation of $f(-2)$ is ‍$0$ because the graph of ‍$y=f(x)$ seems to pass through the point ‍$(-2, 0)$.

   
   

   Evaluating $g({f(-2)})$

   We now know that $g({f(-2)}) \approx g({0})$.

   The best approximation of $g(0)$ is ‍$-3$ because the graph of ‍$y= g(x)$ seems to pass through the point $(0, -3)$.

   
   

   The answer:

   $g(f(-2)) \approx -3$

   ---

2. Strategy

   The composition operator ‍$\circ$ means that we take the output from the second function, ‍‍$g$, and input it into the first function, ‍‍$h$.

   So, the expression ‍‍$(h\circ g) (0)$ is equivalent to‍ $h(g(0))$.
   

   To evaluate ‍$h({g(0)})$, let's first evaluate ‍‍${g(0)}$. Then we'll input that result into ‍$h$ to find the output value.

   From the first table, we see that ‍$g(0)=3$.

    

   We now know that ‍‍$h({g(0)}) = h({3})$. 

   From the second table we see that ‍$h(3)=5$. So ‍$h(g(0))=h(3)=5$.

   
   

   The answer:
   

   ‍$(h\circ g)(0)= {5}$
   

   ---

3. Strategy

   We take the output from the inner function, ‍‍$f$, and input it into the outer function, ‍‍$h$.

   To evaluate ‍$h({f(-2)})$, let's first evaluate ‍‍${f(-2)}$. Then we'll input that result into ‍$h$ to find the output value.

   From the first table, we see that ‍$f(-2)=-2$.

    

   We now know that ‍‍$h({f(-2)}) = h({-2})$. 

   From the second table we see that ‍$h(-2)=2$. So ‍$h(f(-2))=h(-2)=2$.

   
   

   The answer:
   

   ‍$h(f(-2))= {2}$
   

   ---

4. Strategy
   

   The composition operator $\circ$ means that we take the output from the second function, $g$, and input it into the first function, $f$.
   

   To evaluate $f({g(3)})$, let's first evaluate ‍${g(3)}$. Then we'll input that result into $f$ to find the output value.

   Evaluating ${g(3)}$

   The best approximation of $g(3)$ is ‍$-2$ because the graph of ‍$y=g(x)$ seems to pass through the point ‍$(3, -2)$.

   
   

   Evaluating $f({g(3)})$

   We now know that $f({g(3)}) \approx f({-2})$.

   The best approximation of $f(-2)$ is ‍$6$ because the graph of ‍$y= f(x)$ seems to pass through the point $(-2, 6)$.

   
   

   The answer:

   $(f\circ g) (3) \approx 6$