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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #171 : Functions

f(x)=x^2-3 and g(x)=2x+3 . Find $(f\circ g)(x)$ .

Possible Answers:

$(f\circ g)(x)=2x^2+12x+12$

$(f\circ g)(x)=4x^2+12x+6$

$(f\circ g)(x)=2x^2-6$

$(f\circ g)(x)=4x^2-12x+9$

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

Correct answer:

$(f\circ g)(x)=4x^2+12x+6$

Explanation:

${\color{Blue} f(x)=x^2-3}$ and ${\color{Red} g(x)=2x+3}$ .

To find $(f\circ g)(x)$ we plug in the function g(x) everywhere there is a variable in the function f(x) .

This is the composition of "f of g of x".

$(f\circ g)(x)=f(g(x))={\color{Blue} (}{\color{Red} 2x+3}{\color{Blue} )}{\color{Blue} ^2-3}$

Foil the square and simplify:

$(f\circ g)(x)=(4x^2+12x+9)=\mathbf{4x^2+12x+6}$

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Example Question #21 : Composition Of Functions

If f(x)=3x+1 and g(x) = x^3 , what must g(f(-1))+1 be?

Possible Answers:

Correct answer:

Explanation:

Evaluate the composite function g(f(-1)) first.

Solve for f(-1) by substituting into the value for f(x) .

f(-1)=3(-1)+1 = -3+1 = -2

The value of f(-1) will be replaced inside g(f(-1)) , which will become g(-2) .

Evaluate g(-2) .

g(-2) = (-2)^3= (-2)(-2)(-2) = -8

The value of g(f(-1)) is . Add one to this value.

-8+1 = -7

The answer is .

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Example Question #21 : Composition Of Functions

Find f(g(2)) given

f(x)=2x+3 and g(x)=3x-7

Possible Answers:

Correct answer:

Explanation:

To evaluate, first evaluate g(2) and then plug in that answer into f(x) . Thus,

g(2)=3(2)-7=-1

Then, f(-1) is

f(-1)=2(-1)+3=1

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Example Question #1131 : Pre Calculus

Find f(g(1)) given the following.

f(x)=2x+3

g(x)=3x-4

Possible Answers:

Correct answer:

Explanation:

To solve, plug 1 into g and then your answer into f.

Thus,

g(1)=3(1)-4=-1

Plugging in this value into our f function we get the final answer as follows.

f(-1)=2(-1)+3=1

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Example Question #1132 : Pre Calculus

Find g(f(2)) given the following functions:

f(x)=x-6

g(x)=-3x-10

Possible Answers:

Correct answer:

Explanation:

To solve, simply plug in 2 into f and then the result into g.

Thus,

f(2)=2-6=-4

g(-4)=(-3)*(-4)-10=12-10=2

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Example Question #1133 : Pre Calculus

Suppose f(x)=3x^2 and g(x)=x+1 . Find f(g(x)) .

Possible Answers:

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

f(g(x))=3x^2+2x+1

f(g(x))=3x^2+1

f(g(x))=3x^2

f(g(x))=3x^2+6x+3

Correct answer:

f(g(x))=3x^2+6x+3

Explanation:

To find f(g(x)) , you must subsititue g(x) into the function, .

f(g(x))=3(g(x))^2

f(g(x))=3(x+1)^2

f(g(x))=3(x^2+2x+1)

f(g(x))=3x^2+6x+3

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Example Question #23 : Composition Of Functions

If f(x) = x^2 and g(x) = 3x+3 , what is f(g(3))?

Possible Answers:

144

Correct answer:

144

Explanation:

First, we find g(3), or 3(3)+3 =12. Then, find f(12) , or 12^2 = 144.

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Example Question #182 : Functions

Given f(x)=x^2-2x and g(x)=x+2 find f(g(x)) .

Possible Answers:

f(g(x))=x^2+2x+4

f(g(x))=x^2+4x

f(g(x))=x^2+2x

None of these.

f(g(x))=x^2+4x+2

Correct answer:

f(g(x))=x^2+2x

Explanation:

Finding f(g(x)) is the same as plugging in g(x) into f(x) much like one would find f(2){} for a function f(x) .

f(x)=x^2-2x and g(x)=x+2

Insert g(x) into f(x) everywhere there is a variable in f(x):

$\\f(g(x))=(x+2)^2-2(x+2)\\=(x+2)^2-2x-4\\=x^2+4x+4-2x-4\\=\mathbf{x^2+2x}$

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Example Question #1139 : Pre Calculus

We are given the following:

$f(x)=x^2+3\hspace{5mm}$ and g(x)=x+3 .

Find:

$(f\circ g)(x)$

Possible Answers:

$(f\cdot g)(x)=x^2+6x+9$

$(f\cdot g)(x)=2x^2+6x+12$

None of the other answers.

$(f\cdot g)(x)=x^2+3x+6$

$(f\cdot g)(x)=x^2+9x+12$

Correct answer:

None of the other answers.

Explanation:

Let's discuss what the problem is asking us to solve. The expression $(f\cdot g)(x)$ (read as as "f of g of x") is the same as f(g(x)) . In other words, we need to substitute g(x) into f(x) .

Substitute the equation of g(x) for the variable in the given f(x) function:

$(f\cdot g)(x)=(x+3)^{ 2} {+3}$

Next we need to FOIL the squared term and simplify:

(x+3)^2=(x+3)(x+3)

FOIL means that we multiply terms in the following order: first, outer, inner, and last.

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

Outer: $x\cdot 3=3x$

Inner: $3\cdot x=3x$

Last: $3\cdot 3=9$

When we combine like terms, we get the following:

x^2+6x+9

Substitute this back into the equation and continue to simplify.

$(f\cdot g)(x)=x^2+6x+9+3$

$(f\cdot g)(x)=x^2+6x+12$

None of the answers are correct.

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Example Question #1133 : Pre Calculus

Simplify

$\frac{9^{\frac{2}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}$

Possible Answers:

$\frac{9}{2}$

$\frac{9}{4}$

$\frac{27}{2}$

$\frac{3}{4}$

Correct answer:

$\frac{3}{4}$

Explanation:

$\frac{9^{\frac{2}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}=\frac{(3^{2})^{\frac{2}{3}}\cdot3^{\frac{5}{3}}}{6^{2}}=\frac{3^{\frac{4}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}$

$=\frac{3^\frac{9}{3}}{6^2}=\frac{3^3}{6^2}=\frac{27}{36}=\frac{3}{4}$ .

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #171 : Functions

Example Question #21 : Composition Of Functions

Example Question #21 : Composition Of Functions

Example Question #1131 : Pre Calculus

Example Question #1132 : Pre Calculus

Example Question #1133 : Pre Calculus

Example Question #23 : Composition Of Functions

Example Question #182 : Functions

Example Question #1139 : Pre Calculus

Example Question #1133 : Pre Calculus

All Precalculus Resources

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