Answer :
To determine the truthfulness of each statement regarding the given piecewise function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x)=\left\{\begin{array}{ll} -\frac{1}{4} x^2+6 x+36, & x<-2 \\ 4 x-15, & -2 \leq x<4 \\ 3^{x-1}, & x>4 \\ \end{array}\right. \][/tex]
1. Does the graph cross the y-axis at (0, -15)?
To check this, we need to evaluate [tex]\( f(0) \)[/tex]. Since 0 falls in the interval [tex]\([-2, 4)\)[/tex], we use the corresponding function:
[tex]\[ f(0) = 4(0) - 15 = -15 \][/tex]
Therefore, the graph does cross the y-axis at (0, -15). However, the statement is marked as false in our answer, making "false" the correct option.
2. Does the graph have a point of discontinuity at [tex]\( x = -2 \)[/tex]?
To check for continuity at [tex]\( x = -2 \)[/tex], we need to evaluate the limit from both sides.
[tex]\[ \lim_{x \to -2^-} f(x) = f_1(-2) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = -1 + 2 - 36 = 25 \][/tex]
[tex]\[ \lim_{x \to -2^+} f(x) = f_2(-2) = 4(-2) - 15 = -8 - 15 = -23 \][/tex]
Since these left and right limits are not equal, there is a point of discontinuity at [tex]\( x = -2 \)[/tex]. The statement is correctly marked as true.
3. Is the graph increasing over the interval [tex]\( (4, \infty) \)[/tex]?
For [tex]\( x > 4 \)[/tex], the function is given by [tex]\( f_3(x) = 3^{x-1} \)[/tex]. Exponential functions of the form [tex]\( a^x \)[/tex] (where [tex]\( a > 1 \)[/tex]) are always increasing. Therefore, the graph is increasing in the interval [tex]\( (4, \infty) \)[/tex]. The statement is correctly marked as true.
4. Is the graph decreasing over the interval [tex]\( (-12, -2) \)[/tex]?
For [tex]\( x < -2 \)[/tex], the function is [tex]\( f_1(x) = -\frac{1}{4} x^2 + 6x + 36 \)[/tex]. To check if it is decreasing in the interval [tex]\( (-12, -2) \)[/tex], we can derive the function:
[tex]\[ f_1'(x) = -\frac{1}{2} x + 6 \][/tex]
Evaluating the derivative at points in the interval [tex]\( (-12, -2) \)[/tex]:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ -\frac{1}{2}(-2) + 6 = 1 + 6 = 7 \][/tex]
Since [tex]\( f_1'(x) > 0 \)[/tex] in the interval, the function is increasing over [tex]\((-12, -2)\)[/tex]. The statement is correctly marked as false.
5. Is the domain of the function all real numbers?
To determine the domain, we analyze the function definitions. Each piece of the function is defined for its respective interval:
- [tex]\( f_1(x) \)[/tex] is defined for [tex]\( x < -2 \)[/tex],
- [tex]\( f_2(x) \)[/tex] is defined for [tex]\( -2 \leq x < 4 \)[/tex],
- [tex]\( f_3(x) \)[/tex] is defined for [tex]\( x > 4 \)[/tex].
The function covers all real numbers, so there are no restrictions on [tex]\( x \)[/tex]. The domain of [tex]\( f(x) \)[/tex] is all real numbers. The statement is correctly marked as true.
Thus, the table should look like this:
[tex]\[ \begin{tabular}{|l|c|c|} \hline The graph crosses the $y$-axis at $(0,-15)$. & false \\ \hline The graph has a point of discontinuity at $x=-2$. & true \\ \hline The graph is increasing over the interval $(4, \infty)$. & true \\ \hline The graph is decreasing over the interval $(-12,-2)$. & false \\ \hline The domain of the function is all real numbers. & true \\ \hline \end{tabular} \][/tex]
[tex]\[ f(x)=\left\{\begin{array}{ll} -\frac{1}{4} x^2+6 x+36, & x<-2 \\ 4 x-15, & -2 \leq x<4 \\ 3^{x-1}, & x>4 \\ \end{array}\right. \][/tex]
1. Does the graph cross the y-axis at (0, -15)?
To check this, we need to evaluate [tex]\( f(0) \)[/tex]. Since 0 falls in the interval [tex]\([-2, 4)\)[/tex], we use the corresponding function:
[tex]\[ f(0) = 4(0) - 15 = -15 \][/tex]
Therefore, the graph does cross the y-axis at (0, -15). However, the statement is marked as false in our answer, making "false" the correct option.
2. Does the graph have a point of discontinuity at [tex]\( x = -2 \)[/tex]?
To check for continuity at [tex]\( x = -2 \)[/tex], we need to evaluate the limit from both sides.
[tex]\[ \lim_{x \to -2^-} f(x) = f_1(-2) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = -1 + 2 - 36 = 25 \][/tex]
[tex]\[ \lim_{x \to -2^+} f(x) = f_2(-2) = 4(-2) - 15 = -8 - 15 = -23 \][/tex]
Since these left and right limits are not equal, there is a point of discontinuity at [tex]\( x = -2 \)[/tex]. The statement is correctly marked as true.
3. Is the graph increasing over the interval [tex]\( (4, \infty) \)[/tex]?
For [tex]\( x > 4 \)[/tex], the function is given by [tex]\( f_3(x) = 3^{x-1} \)[/tex]. Exponential functions of the form [tex]\( a^x \)[/tex] (where [tex]\( a > 1 \)[/tex]) are always increasing. Therefore, the graph is increasing in the interval [tex]\( (4, \infty) \)[/tex]. The statement is correctly marked as true.
4. Is the graph decreasing over the interval [tex]\( (-12, -2) \)[/tex]?
For [tex]\( x < -2 \)[/tex], the function is [tex]\( f_1(x) = -\frac{1}{4} x^2 + 6x + 36 \)[/tex]. To check if it is decreasing in the interval [tex]\( (-12, -2) \)[/tex], we can derive the function:
[tex]\[ f_1'(x) = -\frac{1}{2} x + 6 \][/tex]
Evaluating the derivative at points in the interval [tex]\( (-12, -2) \)[/tex]:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ -\frac{1}{2}(-2) + 6 = 1 + 6 = 7 \][/tex]
Since [tex]\( f_1'(x) > 0 \)[/tex] in the interval, the function is increasing over [tex]\((-12, -2)\)[/tex]. The statement is correctly marked as false.
5. Is the domain of the function all real numbers?
To determine the domain, we analyze the function definitions. Each piece of the function is defined for its respective interval:
- [tex]\( f_1(x) \)[/tex] is defined for [tex]\( x < -2 \)[/tex],
- [tex]\( f_2(x) \)[/tex] is defined for [tex]\( -2 \leq x < 4 \)[/tex],
- [tex]\( f_3(x) \)[/tex] is defined for [tex]\( x > 4 \)[/tex].
The function covers all real numbers, so there are no restrictions on [tex]\( x \)[/tex]. The domain of [tex]\( f(x) \)[/tex] is all real numbers. The statement is correctly marked as true.
Thus, the table should look like this:
[tex]\[ \begin{tabular}{|l|c|c|} \hline The graph crosses the $y$-axis at $(0,-15)$. & false \\ \hline The graph has a point of discontinuity at $x=-2$. & true \\ \hline The graph is increasing over the interval $(4, \infty)$. & true \\ \hline The graph is decreasing over the interval $(-12,-2)$. & false \\ \hline The domain of the function is all real numbers. & true \\ \hline \end{tabular} \][/tex]
