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Precalculus : Algebra of Functions

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Example Questions

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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Example Question #1131 : Pre Calculus

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

Possible Answers:

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

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

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

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

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

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 #1132 : Pre Calculus

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 #171 : 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 #41 : Algebra Of Functions

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 #42 : Algebra Of Functions

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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Precalculus : Algebra of Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #1131 : Pre Calculus

Example Question #1132 : Pre Calculus

Example Question #171 : Functions

Example Question #41 : Algebra Of Functions

Example Question #42 : Algebra Of Functions

All Precalculus Resources

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