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Precalculus : Composition of Functions

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Example Questions

Example Question #11 : Composition Of Functions

Let

f(x)=e^x

g(x)=x-4

Determine $f\circ g \:(4)$ .

Possible Answers:

e^4

Correct answer:

Explanation:

To find the composite function we start from the most inner portion of the expression and work our way out.

g(4)=4-4=0

$f\circ g \:(4) = f[g(4)]=f(0)=e^0=1$

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Example Question #12 : Composition Of Functions

Let

f(x)=x+4

g(x)=x^2

Determine

$f\circ g$ .

Possible Answers:

x^2+4

x^2+16

(x+4)^2

x^2

Correct answer:

x^2+4

Explanation:

The composite funtion means to replace every entry x in f(x) with the entire function g(x).

$f\circ g= f[g(x)]=(x^2)+4=x^2+4$

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Example Question #11 : Composition Of Functions

For f(x)=x^2 , $g(x)=\frac{x}{3}$ , and $h(x)=3\sqrt{x+1}$ , determine f[g(h(x))] .

Possible Answers:

3x+9

x^2 + 3

$\frac{x+1}{3}$

x+1

$\sqrt{\left(\frac{x}{3}\right)^2 +1}$

Correct answer:

x+1

Explanation:

Working inside out, first do g(h(x)) .

This is,

$\frac{3 \sqrt{x+1}}{3}= \sqrt{x+1}$ .

Now we will do $f(\sqrt{x+1})$ .

This is $(\sqrt{x+1})^2 = x+1$

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Example Question #14 : Composition Of Functions

For $k(x)=ln(x), f(x)=e^{2x}, j(x)=x^2-3$ , write a function for k[f(j(x))] .

Possible Answers:

$e^{(ln^2(x)-3)}$

4x-3

2x^2 -6

4x-ln(3)

4x^2 - 3

Correct answer:

2x^2 -6

Explanation:

Working from the inside out, first we will find a function for f(j(x)) .

This is:

$e^{2(x^2 -3)}$ , which we can simplify slightly to $e^{2x^2 -6}$ .

Now we will plug this new function into the function k:

$ln(e^{2x^2-6})$ .

Since ln is the inverse of e to any power, this simplifies to 2x^2 - 6 .

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Example Question #15 : Composition Of Functions

Find $g\cdot f$ given the following equations

g(x)=2x+3

f(x)=3x+5

Possible Answers:

6x+14

6x^2+19x+15

6x+8

6x+13

Correct answer:

6x+13

Explanation:

To find $g\cdot f$ simply substiute f(x) for every x in g(x) and solve.

g(f(x))=2(3x+5)+3

g(f(x))=6x+13

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Example Question #16 : Composition Of Functions

If f(x)=1-x^2 and g(x)=sin(x) , find $(f\circ g)(x)$ .

Possible Answers:

$\sin (1-x^2)$

$1-\cos^2(x)$

$(1-\sin (x))^2$

$\sin^2(1-x^2)$

$\cos^2(x)$

Correct answer:

$\cos^2(x)$

Explanation:

First, make sure that g $\subseteq$ f (range of g is a subset of the domain of f).

Since the g: $\left \{ y|-1\leq y\leq 1\right \}$ and f: $\left \{ x|x\epsilon \mathbb{R}\right \}$ , g $\subseteq$ f and $(f\circ g)(x)$ exists.

Plug in the output of , which is $\sin (x)$ , as the input of .

Thus,

$(f\circ g)(x)=1-(\sin (x))^2=1-\sin^2(x)=\cos^2 (x)$

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Example Question #17 : Composition Of Functions

Find $(g\cdot f)(x)$ and evaluate at x=5 .

g(x)=x^2-x

f(x)=x-5

Possible Answers:

$(g\cdot f)(x)=x^2-11x+20$

$(g\cdot f)(5)=10$

$(g\cdot f)(x)=x^2-11x+30$

$(g\cdot f)(5)=0$

$(g\cdot f)(x)=x^2-x-5$

$(g\cdot f)(5)=15$

$(g\cdot f)(x)=x^2-x-5$

$(g\cdot f)(5)=10$

$(g\cdot f)(x)=x^2-x-5$

$(g\cdot f)(5)=-15$

Correct answer:

$(g\cdot f)(x)=x^2-11x+30$

$(g\cdot f)(5)=0$

Explanation:

$(g\cdot f)(x)$

$g(x)={\color{Red} x^2-x}$

f(x)=x-5

"G of F of X" means substitute f(x) for the variable in g(x).

$(g\cdot f)(x)=(x-5){\color{Red} ^2}-(x-5)$

Foil the squared term and simplify:

(x-5)^2=(x-5)(x-5)

First: $x\cdot x=x^2$

Outer: $x\cdot -5=-5x$

Inner: $-5 \cdot x= -5x$

Last: $-5 \cdot -5=25$

$(g\cdot f)(x)=x^2-10x+25-x+5=x^2-11x+30$

So $(g\cdot f)(x)={\color{Blue} x^2-11x+30}$

Now evaluate the composite function at the indicated value of x:

$(g\cdot f)(5)=5^2-11(5)+30={\color{Blue} 0}$

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Example Question #14 : Composition Of Functions

Find g(h(x)) if g(x)= x+1 and h(x)= 2x+1 .

Possible Answers:

2x^2+3x+1

3x+3

2x+2

3x+2

2x+3

Correct answer:

2x+2

Explanation:

Replace and substitute the value of h(x) into g(x) so that we are finding g(h(x)) .

g(h(x))= g(2x+1) = (2x+1)+1 = 2x+2

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Example Question #15 : Composition Of Functions

Given h(x) and g(x) , find 3h(g(6)) .

h(x)=4x^3-2^x

g(x)=4x-22

Possible Answers:

Correct answer:

Explanation:

Given h(x) and g(x) , find 3h(g(6)) .

h(x)=4x^3-2^x

g(x)=4x-22

Begin by breaking this into steps. I will begin by computing the g(6) step, because that will make the late steps much more manageable.

g(6)=4(6)-22=24-22=2

Next, take our answer to g(6) and plug it into h(x) .

h(2)=4(2)^3-2^2=32-4=28

So we are close to our final answer, but we still need to multiply by 3.

3h(g(6))=3*28=84

Making our answer 84.

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Example Question #16 : Composition Of Functions

Given $f(x)=\sqrt{x-2}$ and $g(x)=x^2-2\sqrt{x}$ , find $(g\circ f)(x)$ .

Possible Answers:

$(g\circ f)(x)=x-2-2\sqrt[4]{x-2}$

$(g\circ f)(x)=x-2-2\sqrt{x-2}$

$(g\circ f)(x)=-3x-2$

$(g\circ f)(x)=3x-2$

None of the other answers.

Correct answer:

$(g\circ f)(x)=x-2-2\sqrt[4]{x-2}$

Explanation:

$(g\circ f)(x)= g(f(x))$ and is read as "g of f of x" and is equivalent to plugging the function f(x) into the variables in the function g(x).

${\color{Red} f(x)=\sqrt{x-2}}$

${\color{Blue} g(x)=x^2-2\sqrt{x}}$

$(g\circ f)(x)= g(f(x))={\color{Blue}{\color{Red} \sqrt{x-2}}^2-2\sqrt{{\color{Red} \sqrt{x-2}}}}=\mathbf{x-2-2\sqrt[4]{x-2}}$

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Precalculus : Composition of Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Composition Of Functions

Example Question #12 : Composition Of Functions

Example Question #11 : Composition Of Functions

Example Question #14 : Composition Of Functions

Example Question #15 : Composition Of Functions

Example Question #16 : Composition Of Functions

Example Question #17 : Composition Of Functions

Example Question #14 : Composition Of Functions

Example Question #15 : Composition Of Functions

Example Question #16 : Composition Of Functions

All Precalculus Resources

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