Creating and Evaluating Composite Functions Practice

Vadim is trying to evaluate a composite function. That means that he should take the output from the inner function, $f$, and input it into the outer function, $g$.

Let's check each step to learn from his mistake.

Step 1

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

So, the expression $(g\circ f) (-4)$ is equivalent to $g(f(-4))$, not to $f(g(-4))$.

Vadim's mistake is that $(g \circ f)(-4) = g(f(-4))$, not $f(g(-4))$.

Strategy

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

So, the expression $(g\circ h) (-2)$ is equivalent to $g(h(-2))$.

To evaluate $g( {h(-2)})$, let's first evaluate ${h(-2)}$. Then we'll substitute that result into $g$.

${-2} \rightarrow \boxed{h} \rightarrow {h(-2)} \rightarrow \boxed{g} \rightarrow g( {h(-2)})$

Evaluating $h( {-2})$

$\begin{aligned}h(x)&=\left(\dfrac{1}{2}\right)^x\\\\h( {-2})&=\left(\dfrac{1}{2}\right)^{ {-2}}\\\\&= {4}\end{aligned}$

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

We now know ${h(-2)} = {4}$.

$\begin{aligned}g(x)&=-20-3x\\\\g( { {h(-2)}})&=-20-3( {h(-2)})\\\\g( { {4}})&=-20-3( {4})\\\\&=-20-12\\\\&=-32\end{aligned}$

The answer:

$(g\circ h)(-2) =-32$

Strategy

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

To evaluate $f( {h(-5)})$, let's first evaluate ${h(-5)}$. Then we'll substitute that result into $f$ to find our answer.

${-5} \rightarrow \boxed{h} \rightarrow {h(-5)} \rightarrow \boxed{f} \rightarrow f( {h(-5)})$

Evaluating $h( {-5})$

$\begin{aligned} h(t)&=-2t\\\\h( {-5})&=-2( {-5})\\\\&= {10}\end{aligned}$

Evaluating $f( {h(-5)})$

We now know that ${h(-5)} = {10}$.

$\begin{aligned} f(t)&=\dfrac{t}{t-8}\\\\f( {h(-5)})&=\dfrac{ {h(-5)}}{ {h(-5)}-8}\\\\f({ {10}})&=\dfrac{ {10}}{ {10}-8} \\\\&=\dfrac{10}{2}\\\\&=5\end{aligned}$

The answer:

$f(h(-5)) = 5$

Susan is trying to evaluate a composite function. That means that she should take the output from the inner function, $g$, and input it into the outer function, $f$.

Let's check each step to learn from her mistake.

Step 1

To evaluate $g(-15)$, we substitute $-15$ each place where $a$ appears in the function.

$\begin{aligned} g(a)&=\dfrac{2}{3}a\\\\g( {-15})&=\dfrac{2}{3}( {-15})\\\\&=-\dfrac{30}{3}\\\\&= {-10}\end{aligned}$

Susan evaluated $g(-15)$ correctly.

Step 2

To evaluate $f(-15)$, we substitute $-15$ each place where $a$ appears in the function.

$\begin{aligned} f(a)&=-4(a+8)\\\\f({ {-15}})&=-4( {-15}+8) \\\\&=-4(-7)\\\\&=28\end{aligned}$

Susan evaluated $f(-15)$ correctly.

Step 3

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

So, the expression $(f\circ g) (-15)$ is equivalent to $f(g(-15))$, not to $f(-15)\cdot g(-15)$.

Susan's mistake is that $(f \circ g)(-15) = f(g(-15))$, not $f(-15) \cdot g(-15)$.