SAT Math - Graphing Piecewise and Recusive Functions

Define a function as follows:

$g(x) = \left{\begin{matrix} \sin \frac{x}{2} -1 &x\le - \pi \ \cos 2x & -\pi < x \le 0 \ 1-\tan \frac{x }{2}&0< x < \pi \ 0 & x\ge \pi \end{matrix}\right.$

Give the -intercept of the graph of the function.

The graph does not have a -intercept.

Explanation

To find the -intercept, evaluate using the definition of on the interval that includes the value 0. Since

$g(x)= \cos 2x$

on the interval $(-\pi, 0]$ ,

evaluate:

$g(x)= \cos 2 \cdot 0 = \cos 0 = 1$

The -intercept is .

Define a function as follows:

$g(x) = \left{\begin{matrix} x^{2}-2x+7 &x\le -1 \ 9-x & -1 < x \le 0 \ x+8 &0< x \le 1 \ x^{2}+3x+ 6& x> 1 \end{matrix}\right.$

At which of the following values of is the graph of discontinuous?

II)

III)

II and III only

I and II only

I and III only

All of I, II, and III

None of I, II, and III

Explanation

To determine whether is continuous at , we examine the definitions of on both sides of , and evaluate both for :

$x^{2}-2x+7$ evaluated for :

$(-1)^{2}-2(-1)+7= 1+2+7 = 10$

evaluated for :

Since the values coincide, the graph of is continuous at .

We do the same thing with the other two boundary values 0 and 1:

evaluated for :

Since the values do not coincide, the graph of is discontinuous at .

evaluated for :

$x^{2}+3x+ 6$ evaluate for :

$1^{2}+3\cdot 1+ 6 = 1+ 3 + 6 = 10$

Since the values do not coincide, the graph of is discontinuous at .

II and III only is the correct response.

Define function as follows:

$g(x) = \left{\begin{matrix} x^{2}-2x+7 &x\le -1 \ 9-x & -1 < x \le 0 \ x+8 &0< x \le 1 \ x^{2}+3x+ 6& x> 1 \end{matrix}\right.$

Give the -intercept of the graph of the function.

The graph does not have a -intercept.

Explanation

To find the -intercept, evaluate using the definition of on the interval that includes the value 0. Since

on the interval ,

evaluate:

The -intercept is .

Define a function as follows:

$g(x) = \left{\begin{matrix} x^{2}-6x-7 &x\le -5 \ 6-x & -5 < x \le 0 \ x+6 &0< x \le 5\ x^{2}-x- 6& x> 5 \end{matrix}\right.$

How many -intercept(s) does the graph of have?

None

One

Two

Three

Four

Explanation

To find the -coordinates of possible -intercepts, set each of the expressions in the definition equal to 0, making sure that the solution is on the interval on which is so defined.

$g(x)= x^{2}-6x-7$ on the interval $(-\infty, -5]$

$x^{2}-6x-7 = 0$

However, neither value is in the interval $(-\infty, -5]$ , so neither is an -intercept.

on the interval

However, this value is not in the interval , so this is not an -intercept.

on the interval

However, this value is not in the interval , so this is not an -intercept.

$g(x)= x^{2}-x- 6$ on the interval $(5, \infty)$

$x^{2}-x- 6 = 0$

However, neither value is in the interval $(5, \infty)$ , so neither is an -intercept.

The graph of has no -intercepts.

Define a function as follows:

$g(x) = \left{\begin{matrix} 6-x &x\le -5 \ x^{2}-6x-7 & -5 < x \le 0 \x^{2}-x- 6 &0< x \le 5\x+6 & x> 5 \end{matrix}\right.$

How many -intercept(s) does the graph of have?

Two

One

None

Three

Four

Explanation

To find the -coordinates of possible -intercepts, set each of the expressions in the definition equal to 0, making sure that the solution is on the interval on which is so defined.

on the interval $(-\infty, -5]$

However, this value is not in the interval , so this is not an -intercept.

$g(x)= x^{2}-6x-7$ on the interval

$x^{2}-6x-7 = 0$

is on the interval , so is an -intercept.

$g(x)= x^{2}-x- 6$ on the interval

$x^{2}-x- 6 = 0$

is on the interval , so is an -intercept.

on the interval $(5, \infty)$

However, this value is not in the interval $(5, \infty)$ , so this is not an -intercept.

The graph has two -intercepts, and .

Define a function as follows:

$g(x) = \left{\begin{matrix} \sin \frac{x}{2} -1 &x\le - \pi \ \cos 2x & -\pi < x \le 0 \ 1-\tan \frac{x }{2}&0< x < \pi \ 0 & x\ge \pi \end{matrix}\right.$