Creating and Evaluating Composite Functions Practice II

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:

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

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:

‍$(h\circ g)(0)= {5}$

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:

‍$h(f(-2))= {2}$

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:

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