Creating and Evaluating Composite Functions Practice

1. Given:

   $\begin{aligned}&f(b)=\sqrt[3]{2b}+1\\\\&g(b)=28-b\end{aligned}$
   

   Vadim tried to evaluate $(g\circ f)(-4)$, but he made a mistake. Here is his work.
    
   $\begin{array}{lrl}\text{Step 1}&(g\circ f)(-4) &=f(g(-4))\\\\\text{Step 2}&g({-4})&=28-({-4})\\\\&&={32}\\\\\text{Step 3}&f(g(-4))&=f(32)\\\\&&=\sqrt[3]{2({32})}+1 \\\\&&=\sqrt[3]{64}+1\\\\&&=5\end{array}$
   

   What is the mistake in Vadim's work?

   Choose 1 answer:

   1. $(g \circ f)(-4) = g(f(-4))$, not $f(g(-4))$.
   2. $g(-4)=24$, not  $32$. 
   3. $f(32)=9$, not $5$.

   ---

2. $\begin{aligned}&g(x)=-20-3x\\\\&h(x)=\left(\dfrac{1}{2}\right)^x\end{aligned}$

   Evaluate.
   

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

   ---

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

   Evaluate.
   

   $f(h(-5))=$

   ---

4. Given:

   $\begin{aligned}&f(a)=-4(a+8)\\\\&g(a)=\dfrac{2}{3}a\end{aligned}$

   Susan tried to evaluate $(f\circ g)(-15)$, but she made a mistake. Here is her work.

   $\begin{array}{lrl}\text{Step 1}&g(-15) &= \dfrac{2}{3}(-15)\\\\&&=-10\\\\\text{Step 2}&f(-15) &= -4(-15+8)\\\\&&= 28\\\\\text{Step 3}&(f \circ g)(-15) &=28 \cdot -10\\\\&&=-280\end{array}$

   What is the mistake in Susan's work?

   Choose 1 answer:

   1. $f(-15)=68$, not $28$.
   2. $(f \circ g)(-15) = f(g(-15))$, not  $f(-15) \cdot g(-15)$.
   3. $(f \circ g)(-15) = g(f(-15))$, not $f(-15) \cdot g(-15)$.

Source: Khan Academy, https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:composite/x9e81a4f98389efdf:composing/e/evaluate-composite-functions-from-formulas
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. 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 $f(g(-4))$.

   ---

2. 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:

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

   ---

3. 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:

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

   ---

4. 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 $f(-15) \cdot g(-15)$.