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SAT Math : Algebraic Functions

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Example Questions

Example Question #131 : Algebraic Functions

Function 4

Define to be the function graphed above.

Give the -intercept of the graph of the function , which is defined as

g(x) = f(x+ 3)

Possible Answers:

(0, -6)

$\left ( 0, -2 \frac{1}{2}\right )$

The graph of has no -intercept.

(0, -2)

(0,2)

Correct answer:

$\left ( 0, -2 \frac{1}{2}\right )$

Explanation:

The -intercept of a function is the point at which x = 0 , so we can find this by evaluating g(0) .

g(x) = f(x+ 3)

g(0) = f(0+ 3) = f(3)

As can be seen in the diagram below, $f(3) = -2\frac{1}{2}$ .

Function 4a

The -intercept is $\left ( 0, -2 \frac{1}{2}\right )$ .

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Example Question #131 : Algebraic Functions

Two functions

Define and to be the functions graphed above. Evaluate

$\left (g^{-1} \circ f \right ) (5)$ .

Possible Answers:

$\left (g^{-1} \circ f \right ) (5) = -5$

The expression is not defined.

$\left (g^{-1} \circ f \right ) (5) = -3$

$\left (g^{-1} \circ f \right ) (5) = -4$

$\left (g^{-1} \circ f \right ) (5) = 1$

Correct answer:

The expression is not defined.

Explanation:

It can be seen below that a horizontal line can be drawn through two points of the graph of .

Hlt

fails the Horizontal Line Test, which means that has no inverse. $g^{-1}$ does not exist, so the expression $\left (g^{-1} \circ f \right ) (5)$ is undefined.

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Example Question #2832 : Sat Mathematics

Function 4

Define as the function graphed above. Define function g (x) = 3x - 5 .

Evaluate $\left (g \circ f \right ) (3)$ .

Possible Answers:

$\left (g \circ f \right ) (3) = 4$

3 is not in the domain of $g \circ f$ .

$\left (g \circ f \right ) (3) = -12 \frac{1}{2}$

$\left (g \circ f \right ) (3) = -2 \frac{1}{2}$

$\left (g \circ f \right ) (3) = -2$

Correct answer:

$\left (g \circ f \right ) (3) = -12 \frac{1}{2}$

Explanation:

$\left (g \circ f \right ) (3) = g \left [ f (3)\right ]$ .

As can be seen in the diagram below, $f(3) = -2\frac{1}{2} = -\frac{5}{2}$ . Function 4a

Therefore,

$\left (g \circ f \right ) (3) = g \left ( - \frac{5}{2}\right )$

g (x) = 3x - 5 , so

$g \left ( - \frac{5}{2}\right ) = 3\left ( - \frac{5}{2}\right ) -5$

$= - \frac{15}{2} -5$

$= - 7 \frac{1}{2}-5$

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

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Example Question #132 : Algebraic Functions

Two functions

Define and to be the functions graphed above.

Evaluate $(f \circ g ) (4)$

Possible Answers:

$\left (f \circ g \right ) (4) = -3$

$\left (f \circ g \right ) (4) = -6$

$\left (f \circ g \right ) (4) = -2$

4 is not in the domain of $f \circ g$ .

$\left (f \circ g \right ) (4) = -3 \frac{1}{6}$

Correct answer:

$\left (f \circ g \right ) (4) = -2$

Explanation:

$\left (f \circ g \right ) (4) = f \left [ g (4)\right ]$ .

From the diagram below, it can be seen that g(4) = -4

Two functions g 1

Therefore, $\left (f \circ g \right ) (4) = f (-4)$ .

From the diagram below, it can be seen that

f(-4) = -2 .

Two functions f

Therefore, the correct response is that $\left (f \circ g \right ) (4) = -2$ .

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Example Question #133 : Algebraic Functions

Two functions

Define and to be the functions graphed above. Evaluate

$\left (f^{-1} \circ g \right ) (4)$

Possible Answers:

$\left (f ^{-1} \circ g \right ) (4) = -1$

$\left (f ^{-1} \circ g \right ) (4) = -20$

$\left (f ^{-1} \circ g \right ) (4) = -8$

$\left (f ^{-1} \circ g \right ) (4) = 2$

$\left (f^{-1} \circ g \right ) (4)$ is undefined.

Correct answer:

$\left (f ^{-1} \circ g \right ) (4) = -8$

Explanation:

$\left (f ^{-1}\circ g \right ) (4) = f^{-1} \left [ g (4)\right ]$ .

From the diagram below, it can be seen that g(4) = -4

Two functions g 1

Therefore, $\left (f ^{-1} \circ g \right ) (4) = f^{-1} (-4)$ .

From the diagram below, it can be seen that

f(-8) = -4 .

Two functions f

so, by definition,

$f ^{-1} (-4)= -8$ .

Therefore, the correct response is that

$\left (f ^{-1} \circ g \right ) (4) = -8$ .

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Example Question #134 : Algebraic Functions

Function 4

Define as the function graphed above. Define function $g(x) = x^{2}- 11$ .

Evaluate $\left (f \circ g \right ) (4)$ .

Possible Answers:

$\left (f \circ g \right ) (4) = -15$

4 is outside the domain of $f \circ g$

$\left (f \circ g \right ) (4) = - \frac{1}{2}$

$\left (f \circ g \right ) (4) = -10$

$\left (f \circ g \right ) (4) = -7$

Correct answer:

$\left (f \circ g \right ) (4) = - \frac{1}{2}$

Explanation:

$\left (f \circ g \right ) (4) = f \left [ g (4)\right ]$

$g(x) = x^{2}- 11$ , so

$g(4) = 4^{2}- 11 = 16-11 = 5$

$\left (f \circ g \right ) (4) = f (5)$ .

From the diagram below, we see that $f(5) = -\frac{1}{2}$ .

Function 4a

The correct response is that $\left (f \circ g \right ) (4) = - \frac{1}{2}$ .

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Example Question #1061 : Algebra

Two functions

Define and to be the functions graphed above. For which of the following values of is the statement

(f-g)(x) = 0

a true statement?

Possible Answers:

x= 0

x= -1

x= -2

x=1

The statement is not correct for any value of .

Correct answer:

x= 0

Explanation:

f(x) - g(x) = 0

f(x)= g(x)

This can be solved by graphing and on the same set of axes and noting their points of intersection:

Two functions together

The graphs of the two functions intersect at the point (0, 1) . Therefore,

f(0) = g(0) , and

(f-g)(0) = 0 .

The correct response is x= 0 .

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Example Question #91 : How To Find F(X)

Function 4

Define to be the function graphed above.

Which of the following is an -intercept of the graph of the function , if is defined as

g(x ) = f(x+ 5) ?

Possible Answers:

(-3, 0)

(-9, 0)

The graph of has no -intercept.

(-7, 0)

$\left ( -4\frac{1}{2}, 0 \right )$

Correct answer:

(-7, 0)

Explanation:

An -intercept of the graph of has as its -coordinate a value such that

g(x) = 0 ,

or, equivalently,

f(x+ 5) = 0

From the diagram, we can see that

$f(-8) = f(-2) = f \left ( 5 \frac{1}{2}\right ) = 0$

Therefore, to find the -intercepts of , set x+5 equal to these numbers; equivalently, subtract 5 from each number. We get that

$g(-13) = g(-7) = g \left ( \frac{1}{2} \right )= 0$

Therefore, the -intercepts of the graph of are the points

(-13, 0) , (-7, 0) , $\left ( \frac{1}{2}, 0 \right )$ .

The correct choice is (-7, 0) .

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Example Question #136 : Algebraic Functions

Function 4

Define to be the function graphed above.

Give the -intercept of the graph of the function , which is defined as

g(x) = f(x-4) .

Possible Answers:

(0,4)

(0,0)

(0,-8)

The graph of has no -intercept.

(0, -2)

Correct answer:

(0,4)

Explanation:

The -intercept of a function is the point at which x = 0 , so we can find this by evaluating g(0) .

g(x) = f(x-4)

g(0) = f(0-4)= f( -4)

As can be seen in the diagram below, f( -4) = 4 .

Function 4a

Therefore, g(0) = 4 , and the correct response is (0,4) .

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Example Question #137 : Algebraic Functions

Define a function as follows:

$g (x) = x ^{2} - 8x + 17$ , where the domain of is the set [0, 9] .

Give the range of .

Possible Answers:

[17, 26]

$[17, \infty )$

[1, 26]

$[1 , \infty )$

[1, 17]

Correct answer:

[17, 26]

Explanation:

This problem can be solved by examining the behavior of the graph of $g (x) = x ^{2} - 8x + 17$ , which is a parabola.

Since the quadratic coefficient is 1, a positive number, its vertex is a minumum. The -coordinate of the vertex can be found by setting a = 1, b= -8 , and calculating:

$x = - \frac{b}{2a} = -\frac{-8}{2 \cdot 1} = 4$

The -coordinate is

$g (4) = 4 ^{2} - 8 (4) + 17 = 16 - 32 + 17 = 1$

The minimum value of is therefore 1, and this occurs at $x = 4 \in [0, 9]$ . This makes 1 the lower bound of the range.

Since the graph of is a parabola, it decreases everywhere for x < 4 and increases everywhere for x > 4 . Therefore, we can evaluate g(0) and g(9) , and choose the higher value as the maximum value on the given domain.

$g (0) = 0 ^{2} - 8 (0)+ 17 = 0 - 0 + 17 = 17$

$g (0) = 9 ^{2} - 8 (9)+ 17 = 81 - 72 + 17 = 26$

We choose 26 as the upper bound of the range.

Therefore, the range of , given the domain restriction, is the set [17, 26] .

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SAT Math : Algebraic Functions

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #131 : Algebraic Functions

Example Question #131 : Algebraic Functions

Example Question #2832 : Sat Mathematics

Example Question #132 : Algebraic Functions

Example Question #133 : Algebraic Functions

Example Question #134 : Algebraic Functions

Example Question #1061 : Algebra

Example Question #91 : How To Find F(X)

Example Question #136 : Algebraic Functions

Example Question #137 : Algebraic Functions

All SAT Math Resources

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