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SAT II Math II : Functions and Graphs

Study concepts, example questions & explanations for SAT II Math II

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All SAT II Math II Resources

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Example Questions

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Example Question #1 : Solving Quadratic Functions

A baseball is thrown straight up with an initial speed of 60 feet per second by a man standing on the roof of a 100-foot high building. The height of the baseball in feet as a function of time in seconds is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest tenth of a second, how long does it take for the baseball to hit the ground?

Possible Answers:

$1.3\textrm{ sec}$

$6.4\textrm{ sec}$

$5.0 \textrm{ sec}$

$3.8\textrm{ sec}$

$100\textrm{ sec}$

Correct answer:

$5.0 \textrm{ sec}$

Explanation:

When the baseball hits the ground, the height is 0, so we set $h\left ( t\right ) = 0$ . and solve for .

$-16t^{2}+60t + 100 = 0$

This can be done using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Set a = -16, b = 60, c = 100 :

$t = \frac{-60 \pm \sqrt{60^{2}-4(-16)100}}{2(-16)}$

$t = \frac{-60 \pm \sqrt{3,600- (- 6,400)}}{-32}$

$t = \frac{-60 \pm \sqrt{3,600+6,400}}{-32}$

$t = \frac{-60 \pm \sqrt{10,000}}{-32}$

$t = \frac{-60 \pm 100}{-32}$

One possible solution:

$t = \frac{-60 + 100}{-32} = \frac{40}{-32} = -1.25$

We throw this out, since time must be positive.

The other:

$t = \frac{-60 - 100}{-32} = \frac{-160}{-32} = 5$

This solution, we keep. The baseball hits the ground in 5 seconds.

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Example Question #1 : Solving Trigonometric Functions

Give the period of the graph of the equation

$f(x) = \frac{2}{9} \sin 8 x$

Possible Answers:

$\frac{9}{2}$

$\frac{2}{9}$

The correct answer is not among the other choices.

$\frac{1}{8}$

Correct answer:

The correct answer is not among the other choices.

Explanation:

The period of the graph of a sine function $f(x)= A \sin Bx$ is $\frac{2 \pi}{B}$ , or $2 \pi \div B$ .

Since B= 8 ,

$2 \pi \div 8 = \frac{\pi }{4}$ .

This answer is not among the given choices.

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Example Question #2 : Solving Trigonometric Functions

If f(x)=2sin(x)+cos(x) , what must $f(\frac{\pi}{4})$ be?

Possible Answers:

$3\sqrt{2}$

$\textup{The answer is not given.}$

$4\sqrt2$

$\frac{3\sqrt2}{2}$

$\frac{\sqrt2}{2}+2$

Correct answer:

$\frac{3\sqrt2}{2}$

Explanation:

Evaluate each trig function at the specified angle.

$sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

$cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

Replace the terms into the function.

$f(\frac{\pi}{4})=2sin(\frac{\pi}{4})+cos(\frac{\pi}{4}) = 2(\frac{\sqrt{2}}{2})+\frac{\sqrt{2}}{2}$

Combine like-terms.

The answer is: $\frac{3\sqrt2}{2}$

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Example Question #1 : Solving Piecewise And Recusive Functions

Define and as follows:

$f(x) = \left\{\begin{matrix} x+ 2,& x < 0 \\ 4-x, & x \geq 0 \end{matrix}\right.$

$g(x) = \left\{\begin{matrix} 4x ,& x < 3 \\ 3x, & x \geq 3 \end{matrix}\right.$

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

Possible Answers:

$(f \circ g) \left ( 3\right ) = 11$

$(f \circ g) \left ( 3\right ) = -8$

$(f \circ g) \left ( 3\right ) = -5$

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

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

Correct answer:

$(f \circ g) \left ( 3\right ) = -5$

Explanation:

$(f \circ g) \left ( x\right ) = f (g(x))$ by definition.

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

g(x) = 3x on the set $[3, \infty)$ , so

$g(3) = 3 \cdot 3 = 9$ .

f(x) = 4-x on the set $[0 , \infty)$ , so

f (g(3)) = f(9) = 4-9 = -5 .

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Example Question #2 : Solving Piecewise And Recusive Functions

Define function as follows:

$f(x) = \left\{\begin{matrix} 2x- 5 &x< 0 \\ -3x & x \ge 0 \end{matrix}\right.$

Give the range of .

Possible Answers:

$(-\infty, 5)$

(-5, 0]

$(-\infty, 0]$

$(-\infty, \infty)$

$(-\infty, -5)$

Correct answer:

$(-\infty, 0]$

Explanation:

The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.

If x < 0 , then f(x) = 2x - 5 . To find the range of on the interval $(-\infty, 0)$ , we note:

x < 0

$2x < 2 \cdot 0$

2x < 0

2x- 5 < 0 - 5

2x- 5 < - 5

f(x)< - 5

The range of this portion of is $(-\infty, -5)$ .

If $x \ge 0$ , then f(x) = -3x . To find the range of on the interval $[0, \infty)$ , we note:

$x \ge 0$

$-3x \le -3 \cdot 0$

$-3x \le 0$

$f(x)\le 0$

The range of this portion of is $(-\infty, 0 ]$

The union of these two sets is $(-\infty, 0 ]$ , so this is the range of over its entire domain.

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Example Question #1 : Solving Piecewise And Recusive Functions

Define function as follows:

$f(x) = \left\{\begin{matrix} x- 7 &x< -1 \\ 6-x & x > 1 \end{matrix}\right.$

Give the range of .

Possible Answers:

$(-\infty, 5)$

$(-\infty, -8)$

$(-\infty, \infty)$

$(-\infty, -8) \cup (5, \infty)$

$( -8 , \infty)$

Correct answer:

$(-\infty, 5)$

Explanation:

The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.

If x < -1 , then f(x)=x-7 .

To find the range of on the interval $(-\infty, -1)$ , we note:

x < -1

x-7 < -1-7

f(x)< -8

The range of on $(-\infty, -1)$ is $(-\infty, -8)$ .

If x > 1 , then f(x)=6-x .

To find the range of on the interval $(1, \infty )$ , we note:

x > 1

-x < - 1

6-x < 6- 1

f(x)< 5

The range of on $(1, \infty )$ is $(-\infty, 5)$ .

The range of on its entire domain is the union of these sets, or $(-\infty, 5)$ .

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Example Question #1 : Solving Piecewise And Recusive Functions

Define functions and as follows:

f(x)= 4x+7

$g(x) = \left\{\begin{matrix} x+ 6 &x< 0 \\ x- 3 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left ( 2\frac{1}{3}\right )$ .

Possible Answers:

$22\frac{1}{3}$

$4\frac{1}{3}$

$13\frac{1}{3}$

$40\frac{1}{3}$

Undefined

Correct answer:

$4\frac{1}{3}$

Explanation:

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

First, we evaluate $g \left ( 2\frac{1}{3}\right )$ . Since $2\frac{1}{3} \ge 0$ , the definition of for $x \ge 0$ is used, and

g(x)= x-3

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

Since

f(x)= 4x+7 , then

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

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

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

$= 4 \left ( -\frac{2}{3} \right )+7$

$= -2\frac{2}{3} +7$

$= 4\frac{1}{3}$

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Example Question #5 : Solving Piecewise And Recusive Functions

Define functions and as follows:

$f(x) = \left\{\begin{matrix} 2x &x< -1 \\ 3x & x > 1 \end{matrix}\right.$

$g(x) = \left\{\begin{matrix} x+ 5 &x< -1 \\ x- 2 & x > 1 \end{matrix}\right.$

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

Possible Answers:

Undefined

Correct answer:

Undefined

Explanation:

$(f \circ g)(3) = f(g(3))$

First, evaluate g(3) using the definition of for x > 1 :

g(x)= x-2

g(3)= 3-2 = 1

Therefore,

$(f \circ g)(3) = f(g(3)) = f(1)$

However, x = 1 is not in the domain of .

Therefore, $(f \circ g)(3)$ is an undefined quantity.

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Example Question #6 : Solving Piecewise And Recusive Functions

Define functions and as follows:

$f(x) = \left\{\begin{matrix} 2x &x< -1 \\ 3x & x > 1 \end{matrix}\right.$

$g(x) = \left\{\begin{matrix} x+ 5 &x< -1 \\ x- 2 & x > 1 \end{matrix}\right.$

Evaluate $(g \circ f)(-3)$ .

Possible Answers:

Undefined

Correct answer:

Explanation:

$(g \circ f)(-3)= g(f(-3))$

First, evaluate f(-3) using the definition of for x < -1 :

f(x)= 2x

f(-3)= 2(-3) = -6

Therefore,

$(g \circ f)(-3)= g(f(-3)) = g(-6)$

Evaluate g(-6) using the definition of for x < -1 :

g(x)= x+ 5

g(-6) = -6+ 5 = -1

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Example Question #2 : Solving Piecewise And Recusive Functions

Define functions and as follows:

$f(x) = \left\{\begin{matrix} 2x+ 1 &x< 0 \\ 3x & x \ge 0 \end{matrix}\right.$

$g(x) = \left\{\begin{matrix} x+ 9 &x< 0 \\ x- 2 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left (-1.7\right )$ .

Possible Answers:

-10.1

15.6

-2.4

Undefined

21.9

Correct answer:

21.9

Explanation:

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right )$

First we evaluate g(-1.7) . Since -1.7 < 0 , we use the definition of for the values in the range x < 0 :

g(x)= x + 9

g(-1.7)= -1.7+ 9 = 7.3

Therefore,

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right ) = f (7.3)$

Since $7.3 \ge 0$ , we use the definition of for the range $x \ge 0$ :

f(x)= 3x

$f (7.3) = 3 \cdot 7.3 = 21.9$

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SAT II Math II : Functions and Graphs

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : Solving Quadratic Functions

Example Question #1 : Solving Trigonometric Functions

Example Question #2 : Solving Trigonometric Functions

Example Question #1 : Solving Piecewise And Recusive Functions

Example Question #2 : Solving Piecewise And Recusive Functions

Example Question #1 : Solving Piecewise And Recusive Functions

Example Question #1 : Solving Piecewise And Recusive Functions

Example Question #5 : Solving Piecewise And Recusive Functions

Example Question #6 : Solving Piecewise And Recusive Functions

Example Question #2 : Solving Piecewise And Recusive Functions

All SAT II Math II Resources

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