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Precalculus : Algebra of Functions

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Example Questions

Example Question #4 : Composition Of Functions

If f(x)=x^2-3 , g(x)=6x+4 , and h(x)=x+8 , what is f(g(h(-7))) ?

Possible Answers:

288

Correct answer:

Explanation:

When doing a composition of functions such as this one, you must always remember to start with the innermost parentheses and work backward towards the outside.

So, to begin, we have

h(-7)=(-7)+8=1 .

Now we move outward, getting

g(h(-7))=g(1)=6(1)+4=10 .

Finally, we move outward one more time, getting

f(g(h(-7)))=f(g(1))=f(10)=(10)^2-3=97 .

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Example Question #3 : Composition Of Functions

Find g(h(i(2))) if g(x) = x , h(x) = 3 , and i(x)=x^4 .

Possible Answers:

3+x

48x

Correct answer:

Explanation:

Solve for the value of i(2) .

i(2)= 2^4=16

Solve for the value of h(i(2)) .

h(i(2))=h(16) = 3

Solve for the value g(h(i(2))) .

g(h(i(2)))=g(3) = 3

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Example Question #1 : Composition Of Functions

For the functions f(x)=x+7 and $g(x)=x^{3}$ , evaluate the composite function $f\circ g$

Possible Answers:

$x^{3}+7$

$x^{3}+21x^{2}$

$(x+7)^{3}$

$x^{3}+7x^{2}+14x+21$

$x^{3}+49x^{2}$

Correct answer:

$x^{3}+7$

Explanation:

The composite function notation $f\circ g$ means to swap the function g(x) into f(x) for every value of . Therefore:

$f\circ g$

=f[g(x)]

$=(x^{3})+7$

$=x^{3}+7$

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Example Question #2 : Composition Of Functions

For the functions f(x)=x-7 and g(x)=4x+6 , evaluate the composite function $g\circ f$ .

Possible Answers:

4x-13

3x+13

3x-1

4x-22

4x-1

Correct answer:

4x-22

Explanation:

The composite function notation $g\circ f$ means to swap the function f(x) into g(x) for every value of . Therefore:

$g\circ f$

=g[f(x)]

=4(x-7)+6

=4x-28+6

=4x-22

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Example Question #1 : Composition Of Functions

For the functions $f(x)=x^{2}-4$ and g(x)=2x+3 , evaluate the composite function $f\circ g$ .

Possible Answers:

$2x^{2}-1$

$4x^{2}+12x+9$

$2x^{2}-4$

$4x^{2}+12x+5$

None of the answers listed

Correct answer:

$4x^{2}+12x+5$

Explanation:

The composite function notation $f\circ g$ means to swap the function g(x) into f(x) for every value of . Therefore:

$f\circ g$

=f[g(x)]

$=(2x+3)^{2}-4$

$=4x^{2}+12x+9-4$

$=4x^{2}+12x+5$

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Example Question #1121 : Pre Calculus

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 #1122 : Pre Calculus

Let

f(x)=x+4

g(x)=x^2

Determine

$f\circ g$ .

Possible Answers:

x^2

x^2+16

(x+4)^2

x^2+4

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 #1123 : Pre Calculus

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

Possible Answers:

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

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

x+1

x^2 + 3

3x+9

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 #34 : Algebra 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:

4x-ln(3)

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

4x-3

2x^2 -6

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 #13 : Composition Of Functions

Find $g\cdot f$ given the following equations

g(x)=2x+3

f(x)=3x+5

Possible Answers:

6x+8

6x+14

6x+13

6x^2+19x+15

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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Precalculus : Algebra of Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #4 : Composition Of Functions

Example Question #3 : Composition Of Functions

Example Question #1 : Composition Of Functions

Example Question #2 : Composition Of Functions

Example Question #1 : Composition Of Functions

Example Question #1121 : Pre Calculus

Example Question #1122 : Pre Calculus

Example Question #1123 : Pre Calculus

Example Question #34 : Algebra Of Functions

Example Question #13 : Composition Of Functions

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