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SAT II Math II : Solving Piecewise and Recusive Functions

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Example Questions

Example Question #1 :Solving Piecewise And Recusive Functions

Defineandas 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:

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

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

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

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

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

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 #1 :Solving Piecewise And Recusive Functions

Define functionas follows:

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

Give the range of.

Possible Answers:

$(-\infty, \infty)$

$(-\infty, 5)$

(-5, 0]

$(-\infty, 0]$

$(-\infty, -5)$

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 ofon 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 ofis $(-\infty, -5)$ .

If $x \ge 0$ , then f(x) = -3x . To find the range ofon 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 ofis $(-\infty, 0 ]$

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

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Example Question #3 :Solving Piecewise And Recusive Functions

Define functionas follows:

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

Give the range of.

Possible Answers:

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

$(-\infty, 5)$

$( -8 , \infty)$

$(-\infty, -8)$

$(-\infty, \infty)$

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 ofon the interval $(-\infty, -1)$ , we note:

x < -1

x-7 < -1-7

f(x)< -8

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

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

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

x > 1

-x < - 1

6-x < 6- 1

f(x)< 5

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

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

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Example Question #4 :Solving Piecewise And Recusive Functions

Define functionsandas 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:

$4\frac{1}{3}$

Undefined

$40\frac{1}{3}$

$13\frac{1}{3}$

$22\frac{1}{3}$

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 offor $x \ge 0$ 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 #2 :Solving Piecewise And Recusive Functions

Define functionsandas 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 offor 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 functionsandas 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 offor 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 offor x < -1 :

g(x)= x+ 5

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

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Example Question #7 :Solving Piecewise And Recusive Functions

Define functionsandas 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:

15.6

Undefined

-2.4

21.9

-10.1

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 offor 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 offor the range $x \ge 0$ :

f(x)= 3x

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

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Example Question #8 :Solving Piecewise And Recusive Functions

Define two functions as follows:

$f(x) = \begin{Bmatrix} 2x- 7& \textrm{for } x < 0 \\ & \\ 3x+ 3& \textrm{for } x\ge 0 \end{matrix}$

$g(x) = \begin{Bmatrix} 3x+ 3 & \textrm{for } x < 0 \\ & \\ 2x- 7 & \textrm{for } x\ge 0 \end{matrix}$

Evaluate $(f \circ g) (5)$ .

Possible Answers:

Correct answer:

Explanation:

By definition, $(f \circ g) (5) = f \left [ g (5) \right ]$

First, evaluate g(5) , using the definition of g(x) for nonnegative values of. Substitutingfor 5:

g(x) = 2x-7

$g(5) = 2\cdot 5 -7= 10 -7 = 3$

$(f \circ g) (5) = f \left [ g (5) \right ] = f(3)$ ; evaluate this using the definition of f(x) for nonnegative values of:

f(x) = 3x+ 3

$f(3) = 3 \cdot 3+ 3 = 9 + 3 = 12$

12 is the correct value.

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Example Question #9 :Solving Piecewise And Recusive Functions

Which of the following would be a valid alternative definition for the provided function?

$f(x) = \left | \frac{3}{4} x - \frac{7}{3} \right |$

Possible Answers:

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < -\frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge - \frac{28}{9} \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{3}{4} x - \frac{7}{3} & \textrm{for }x < - \frac{28}{9}\\ \\ - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x \ge - \frac{28}{9} \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{3}{4} x - \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

None of these

Correct answer:

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

Explanation:

The absolute value of an expression | a | is defined as follows:

| a | = a for $a \ge 0$

| a | = -a for a < 0

Therefore,

$f(x) = \left | \frac{3}{4} x - \frac{7}{3} \right | = \frac{3}{4} x - \frac{7}{3}$

if and only if

$\frac{3}{4} x - \frac{7}{3} \ge 0$ .

Solving this condition for:

$\frac{3}{4} x - \frac{7}{3} + \frac{7}{3} \ge 0 + \frac{7}{3}$

$\frac{3}{4} x \ge \frac{7}{3}$

$\frac{4}{3} \cdot \frac{3}{4} x \ge \frac{4}{3} \cdot \frac{7}{3}$

$x \ge \frac{28}{9}$

Therefore, $f(x) = \frac{3}{4} x - \frac{7}{3}$ for $x \ge \frac{28}{9}$ .

Similarly,

$f(x) = - \left (\frac{3}{4} x - \frac{7}{3} \right ) = - \frac{3}{4} x + \frac{7}{3}$ for $x < \frac{28}{9}$ .

The correct response is therefore

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

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SAT II Math II : Solving Piecewise and Recusive Functions

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

All SAT II Math II Resources

Example Questions

Example Question #1 :Solving Piecewise And Recusive Functions

Example Question #1 :Solving Piecewise And Recusive Functions

Example Question #3 :Solving Piecewise And Recusive Functions

Example Question #4 :Solving Piecewise And Recusive Functions

Example Question #2 :Solving Piecewise And Recusive Functions

Example Question #6 :Solving Piecewise And Recusive Functions

Example Question #7 :Solving Piecewise And Recusive Functions

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