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 $5$.

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

   Evaluate.
   

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

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

   Evaluate.
   

   $f(h(-5))=$

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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
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