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SAT Math : Algebra

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Example Question #1091 : Algebra

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

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

Possible Answers:

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

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

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

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

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

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 #1092 : 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- 16

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

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

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

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

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 #1091 : Algebra

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) =x+ 8

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

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

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

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 #121 : How To Find F(X)

Define two functions as follows:

f(x) = 4x- 17

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

Evaluate (f+g)(19) .

Possible Answers:

$36 \frac{1}{2}$

$81\frac{1}{2}$

$5 \frac{1}{9}$

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

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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Example Question #1095 : Algebra

Define two functions as follows:

f(x) = 4x- 17

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

Evaluate (f-g)(19) .

Possible Answers:

$81\frac{1}{2}$

$36 \frac{1}{2}$

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

$5 \frac{1}{9}$

Correct answer:

$36 \frac{1}{2}$

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 subtract:

f(x) = 4x- 17

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

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

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

$(f-g)(19) = f(19)- g(19) = 59 - 22 \frac{1}{2} = 36 \frac{1}{2}$

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Example Question #1096 : Algebra

Define two functions as follows:

f(x) = 4x- 17

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

$(f \circ g) (c) = 10$ .

Evaluate .

Possible Answers:

$2 \frac{3}{4}$

$24 \frac{1}{2}$

$-12 \frac{1}{2}$

None of the other choices gives the correct response.

Correct answer:

$-12 \frac{1}{2}$

Explanation:

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

Replacing g(x) with its definition, we get

$(f \circ g ) (x) = f \left ( \frac{1}{2}x+ 13 \right )$

In the definition of f ( x) , replace with $\frac{1}{2}x+ 13$ and simplify the expression:

f(x) = 4x- 17

$f \left ( \frac{1}{2}x+ 13 \right )= 4 \left ( \frac{1}{2}x+ 13 \right )- 17$

= 2 x+ 52 - 17

= 2 x+ 35

Therefore,

$(f \circ g ) (x) = 2 x+ 35$

$(f \circ g) (c) = 10$ ,

then

2 c+35 = 10

Solve for :

2 c+35-35 = 10 - 35

2 c = -25

$\frac{1}{2} \cdot 2 c = \frac{1}{2} \cdot (-25)$

$c = -12 \frac{1}{2}$

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

Define two functions as follows:

f(x) = 4x- 17

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

$(g \circ f ) (c) = 10$ .

Evaluate .

Possible Answers:

$24 \frac{1}{2}$

None of the other choices gives the correct response.

$2 \frac{3}{4}$

$-12 \frac{1}{2}$

Correct answer:

$2 \frac{3}{4}$

Explanation:

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

Replacing f(x) with its definition, we get

$(g \circ f ) (x) = g ( 4x- 17)$

In the definition of g ( x) , replace with 4x-17 and simplify the expression:

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

$g(4x-17 ) = \frac{1}{2} (4x-17 )+ 13$

$= 2 x-8 \frac{1}{2} + 13$

$= 2 x+4 \frac{1}{2}$

Therefore,

$(g \circ f ) (x) = 2 x+4 \frac{1}{2}$

$(g \circ f ) (c) = 10$ ,

then

$2 c+4 \frac{1}{2} = 10$

Solve for :

$2 c+4 \frac{1}{2} - 4 \frac{1}{2} = 10 - 4 \frac{1}{2}$

$2 c = 5 \frac{1}{2}$

$\frac{1}{2} \cdot 2 c = \frac{1}{2} \cdot 5 \frac{1}{2}$

$c = 2 \frac{3}{4}$

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Example Question #1098 : Algebra

Define two functions as follows:

f(x) = 4x- 17

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

(f-g) (c) = 19

Evaluate .

Possible Answers:

$c = 36 \frac{1}{2}$

$c = 81\frac{1}{2}$

$c = 5 \frac{1}{9}$

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

c =14

Correct answer:

c =14

Explanation:

To obtain the definition of the function (f-g)(x) , subtract the expressions that define the individual functions f(x) and g(x) :

f(x) = 4x- 17

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

$(f+g)(x)= (4x- 17 )- \left ( \frac{1}{2}x+ 13 \right )$

$= 4x- \frac{1}{2}x- 17 -13$

$= \frac{7}{2}x- 30$

(f+g) (c) = 19 , so

$\frac{7}{2}c- 30 = 19$

Solve for ; add 30:

$\frac{7}{2}c- 30 + 30 = 19 + 30$

$\frac{7}{2}c = 49$

Multiply by $\frac{2}{7}$ :

$\frac{2} {7}\cdot \frac{7}{2}c = \frac{2} {7}\cdot 49$

c =14

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Example Question #1099 : Algebra

Define two functions as follows:

f(x) = 4x- 17

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

(f+g) (c) = 19

Evaluate .

Possible Answers:

$c = 5 \frac{1}{9}$

$c = 81\frac{1}{2}$

c =14

$c = 36 \frac{1}{2}$

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

Correct answer:

$c = 5 \frac{1}{9}$

Explanation:

To obtain the definition of the function (f+g)(x) , add the expressions that define the individual functions f(x) and g(x) "

f(x) = 4x- 17

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

$(f+g)(x)= (4x- 17 )+ \left ( \frac{1}{2}x+ 13 \right )$

$= 4x+ \frac{1}{2}x- 17 + 13$

$= \frac{9}{2}x- 4$

(f+g) (c) = 19 , so

$\frac{9}{2}c- 4 = 19$ ;

Solve for ; add 4;

$\frac{9}{2}c- 4+ 4 = 19 + 4$

$\frac{9}{2}c = 23$

Multiply by $\frac{2}{9}$ :

$\frac{2}{9} \cdot \frac{9}{2}c =\frac{2}{9} \cdot 23$

$c =\frac{46}{9} = 5 \frac{1}{9}$

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Example Question #1100 : Algebra

Define f(x) = 7x - 17 , restricting the domain to $x \in [-4, 8]$ .

Give the range of .

Possible Answers:

$(-\infty, \infty)$

$\left [ -147, -63\right ]$

$\left [ -45, 39 \right ]$

$\left [ -147, \infty \right )$

$\left [ -45, \infty \right )$

Correct answer:

$\left [ -45, 39 \right ]$

Explanation:

A function of the form f(x) = m x + b is a linear function and is either constantly increasing or constantly decreasing. Therefore, f(x) = 7x - 17 has its minimum and maximum values at the endpoints of its domain.

We evaluate f(-4) and f(8) by substitution, as follows:

f(x) = 7x - 17

f(-4) = 7 (-4) - 17 = -28 - 17 = -45

f(8) = 7 (8) - 17 = 56 - 17 = 39

The range of the function on the domain to which it is restricted is $\left [ -45, 39 \right ]$ .

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SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1091 : Algebra

Example Question #1092 : Algebra

Example Question #1091 : Algebra

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

Example Question #1095 : Algebra

Example Question #1096 : Algebra

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

Example Question #1098 : Algebra

Example Question #1099 : Algebra

Example Question #1100 : Algebra

All SAT Math Resources

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