Solving Piecewise and Recusive Functions - SAT Subject Test in Math II

Card 0 of 9

Question

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

Answer

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

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

on the set $[3, \infty)$ , so

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

on the set $[0 , \infty)$ , so

.

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Question

Define functions and as follows:

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

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

Answer

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

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

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

Since

, 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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Question

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

Answer

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

First we evaluate . Since , we use the definition of for the values in the range :

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 $x \ge 0$ :

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

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Question

Define function as follows:

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

Give the range of .

Answer

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

If , then . To find the range of on the interval $(-\infty, 0)$ , we note:

$2x < 2 \cdot 0$

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

If $x \ge 0$ , then . 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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Question

Define function as follows:

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

Give the range of .

Answer

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

If , then .

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

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

If , then .

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

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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Question

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

Answer

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

First, evaluate using the definition of for :

Therefore,

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

However, is not in the domain of .

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

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Question

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

Answer

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

First, evaluate using the definition of for :

Therefore,

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

Evaluate using the definition of for :

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Question

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

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

Answer

The absolute value of an expression is defined as follows:

for $a \ge 0$

for

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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Question

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

Answer

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

First, evaluate , using the definition of for nonnegative values of . Substituting for 5:

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

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

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

12 is the correct value.

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