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SAT Math : SAT Mathematics

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Example Questions

Example Question #1083 : Algebra

If $f(x)=\left | x+1\right |$ then what value of will make f(x)=12 true?

Possible Answers:

$\pm11$

11, -13

$\pm13$

Correct answer:

11, -13

Explanation:

We know that $f(x)=\left | x+1\right |$ and f(x)=12 . Just set them equal to each other.

$\left | x+1\right |=12$ Remember to account for negative values.

x+1=12 Subtract on both sides.

x=11

x+1=-12 Subtract on both sides.

x=-13

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Example Question #1083 : Algebra

If f(x)=x+5 and g(x)=x^2 then what value of will make f(g(x))=30 true?

Possible Answers:

$\pm5$

Correct answer:

$\pm5$

Explanation:

We know f(g(x))=30 so we need to apply substitutions to solve for .

f(g(x))=f(x^2)=x^2+5=30 Subtract on both sides.

x^2=25 Take square root on both sides and account for negative values.

$x=\pm5$

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Example Question #1084 : Algebra

If f(x)=x-1, g(x)=x^2+x , then what value of will make f(g(x))=5

Possible Answers:

-3, 2

2,-3

$\pm2, \pm3$

$\pm3$

$\pm2$

Correct answer:

-3, 2

Explanation:

f(x)=x-1, g(x)=x^2+x

We know f(g(x))=5 so let's make the substitution.

f(g(x))=f(x^2+x)=x^2+x-1=5 This is a quadratic so subtract on both sides.

x^2+x-6=0 Factor.

(x+3)(x-2)=0 Solve individually.

x=-3, 2

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Example Question #1085 : Algebra

Define f(x) = 7x - 15 .

How can g(x) be defined so that $(f \circ g) (x) = 14x + 4$ ?

Possible Answers:

$g(x) = \frac{1}{2}x - \frac{ 11}{14}$

g(x)= -7x - 19

$g(x) = \frac{14x+ 4}{7x-15}$

$g(x) = 2x + \frac{ 19}{7}$

$g(x) = \frac{7x-15}{14x+ 4}$

Correct answer:

$g(x) = 2x + \frac{ 19}{7}$

Explanation:

By definition,

$(f \circ g) (x) = f [g(x)]$ ,

f [g(x)] = 14x + 4

f(x) = 7x - 15 ,

it follows that

f [g(x)] =7g(x) - 15 ,

and, substituting,

7g(x) - 15 = 14x + 4

Solving for g(x) by isolating this expression:

7g(x) - 15 + 15 = 14x + 4 + 15

7g(x) = 14x + 19

$7g(x) \div 7 = (14x + 19) \div 7$

$g(x) = \frac{14x + 19}{7}$

$g(x) = 2x + \frac{ 19}{7}$ .

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Example Question #1086 : Algebra

Define $f(x) = x^{2} + 4$ .

How can g(x) be defined so that $(f \circ g) (x) = 6x$ ?

Possible Answers:

$g(x) = \sqrt{ \frac{x - 4}{6}}$

$g(x) = 6 \sqrt{ x - 4}$

$g(x) = \sqrt{ 6x - 4}$

$g(x) = \frac{1}{6} \sqrt{ x - 4}$

$g(x) = -4 + \sqrt{ 6x }$

Correct answer:

$g(x) = \sqrt{ 6x - 4}$

Explanation:

By definition,

$(f \circ g) (x) = f [g(x)]$ ,

f [g(x)] = 6x

$f(x) = x^{2} + 4$ ,

it follows that

$f [g(x)] = [g(x)]^{2} + 4$ ,

and, substituting,

$[g(x)]^{2} + 4 = 6x$

Solving for g(x) by isolating this expression:

$[g(x)]^{2} + 4 = 6x$

$[g(x)]^{2} = 6x - 4$

Taking the square root of both sides:

$g(x) = \pm \sqrt{ 6x - 4}$

Either $g(x) =- \sqrt{ 6x - 4}$ , which is not among the given choices, or $g(x) = \sqrt{ 6x - 4}$ , which is.

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Example Question #1087 : Algebra

Define $f(x) = \sqrt{3x+ 21 }$ .

How can g(x) be defined so that $(f \circ g) (x) = 9x$ ?

Possible Answers:

$g(x) = 9 x^{2} -49$

$g(x) = 27 x^{2} - 21$

$g(x) = 27 x^{2} - 7$

$g(x) = 9 x^{2} -42x +49$

$g(x) = 9 x^{2} + 42x +49$

Correct answer:

$g(x) = 27 x^{2} - 7$

Explanation:

By definition,

$(f \circ g) (x) = f [g(x)]$ ,

f [g(x)] = 9x

$f(x) = \sqrt{3x+ 21}$ ,

it follows that

$f [g(x)] = \sqrt{3 [g(x)]+ 21}$ ,

and, substituting,

$\sqrt{3 [g(x)]+ 21 } = 9x$

Solving for g(x) by isolating this expression:

$\left (\sqrt{3 [g(x)]+ 21} \right ) ^{2}=\left ( 9x \right )^{2}$

Applying the Power of a Product Rule:

$3 [g(x)]+ 21= 9^{2} \cdot x^{2}$

$3 [g(x)]+ 21= 81 x^{2}$

$3 [g(x)]+ 21- 21 = 81 x^{2} - 21$

$3 [g(x)] = 81 x^{2} - 21$

$\frac{3 [g(x)] }{3} = \frac{ 81 x^{2} - 21 }{3}$

$g(x) = 27 x^{2} - 7$

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Example Question #121 : How To Find F(X)

Define $f(x) = \frac{1}{x- 5}$ .

How can g(x) be defined so that $(f \circ g) (x) = x ^{2}$ ?

Possible Answers:

$g(x) = \frac{25x^{2} +10x+ 1}{x^{2} }$

$g(x) = \frac{25x^{2} -10x+ 1}{x^{2} }$

$g(x) = \frac{5 x ^{2} + 1 }{ x ^{2} }$

$g(x) = \frac{25x^{2} - 1}{x^{2} }$

$g(x) = \frac{ x ^{2} + 5 }{ x ^{2} }$

Correct answer:

$g(x) = \frac{5 x ^{2} + 1 }{ x ^{2} }$

Explanation:

By definition,

$(f \circ g) (x) = f [g(x)]$ ,

$f [g(x)] = x^{2}$

$f(x) = \frac{1}{x- 5}$ ,

it follows that

$f [g(x)] = \frac{1}{ g(x)- 5}$ ,

and, substituting,

$\frac{1}{ g(x)- 5} = x ^{2}$

Solving for g(x) by isolating this expression, we first take the reciprocal of both sides:

$g(x)- 5 = \frac{1}{ x ^{2} }$

$g(x)- 5 +5 = \frac{1}{ x ^{2} } + 5$

$g(x) = \frac{1}{ x ^{2} } + \frac{5 x ^{2} }{ x ^{2} }$

$g(x) = \frac{5 x ^{2} + 1 }{ x ^{2} }$

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Example Question #1091 : Algebra

Define $f(x) = x^{2} + 8$ .

How can g(x) be defined so that $(f \circ g) (x) = x^{2} - 8$ ?

Possible Answers:

$g(x) = x^{4}- 16$

$g(x) = \sqrt{x^{2} - 16}$

$g(x) = \sqrt{x - 16}$

g(x) = x- 16

$g(x) = x^{2}- 16$

Correct answer:

$g(x) = \sqrt{x^{2} - 16}$

Explanation:

By definition,

$(f \circ g) (x) = f [g(x)]$ ,

$f [g(x)] = x^{2} - 8$

$f(x) = x^{2} +8$ ,

it follows that

$f [g(x)] =[ g(x) ]^{2} + 8$ ,

and, substituting,

$[ g(x) ]^{2} + 8 = x^{2} - 8$

Solving for g(x) by isolating this expression:

$[ g(x) ]^{2} + 8- 8 = x^{2} - 8 - 8$

$[ g(x) ]^{2} = x^{2} - 16$

Taking the square root of both sides:

$g(x) = \pm \sqrt{x^{2} - 16}$ ,

or, either $g(x) = \sqrt{x^{2} - 16}$ or $g(x) = - \sqrt{x^{2} - 16}$ . The second definition is not among the choices; the first one is, and is the correct response.

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Example Question #164 : Algebraic Functions

Define $f(x) = \frac{1}{x- 4}$ .

How can g(x) be defined so that $(f \circ g) (x) = x+ 4$ ?

Possible Answers:

$g(x) = \frac{8x+ 1}{x}$

$g(x) = \frac{17x+4}{x+ 4 }$

$g(x) = \frac{x+ 8}{x}$

g(x) =x+ 8

$g(x) = \frac{4x+17}{x+ 4 }$

Correct answer:

$g(x) = \frac{4x+17}{x+ 4 }$

Explanation:

By definition,

$(f \circ g) (x) = f [g(x)]$ ,

f [g(x)] = x+ 4

$f(x) = \frac{1}{x- 4}$ ,

it follows that

$f [g(x)] = \frac{1}{ g(x)- 4}$ ,

and, substituting,

$\frac{1}{ g(x)- 4} = x+ 4$

Solving for g(x) by isolating this expression, we first take the reciprocal of both sides:

$g(x)- 4= \frac{1}{x+ 4}$

Now, we can isolate g(x) :

$g(x)- 4+ 4 = \frac{1}{x+ 4}+ 4$

Simplify the expression on the right:

$g(x) = \frac{1}{x+ 4}+ \frac{4(x+4)}{x+ 4 }$

$g(x) = \frac{1}{x+ 4}+ \frac{4x+16}{x+ 4 }$

$g(x) = \frac{4x+17}{x+ 4 }$

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Example Question #1092 : Algebra

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

Evaluate (f+g)(19) .

Possible Answers:

$81\frac{1}{2}$

$5 \frac{1}{9}$

$36 \frac{1}{2}$

$-3\frac{1}{7}$

Correct answer:

$81\frac{1}{2}$

Explanation:

By definition, (f+g)(19) = f(19)+ g(19) ; simply evaluate f(19) and g(19) by substituting 19 for in both definitions, and adding:

f(x) = 4x- 17

$f(19) = 4 \cdot 19 - 17 = 76 - 17 = 59$

$g(x) = \frac{1}{2}x+ 13$

$g(19) = \frac{1}{2} \cdot 19 + 13 = 9\frac{1}{2} + 13 = 22 \frac{1}{2}$

$(f+g)(19) = f(19)+ g(19) = 59 + 22 \frac{1}{2} = 81 \frac{1}{2}$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1083 : Algebra

Example Question #1083 : Algebra

Example Question #1084 : Algebra

Example Question #1085 : Algebra

Example Question #1086 : Algebra

Example Question #1087 : Algebra

Example Question #121 : How To Find F(X)

Example Question #1091 : Algebra

Example Question #164 : Algebraic Functions

Example Question #1092 : Algebra

All SAT Math Resources

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