Answer :
To solve this problem, let's analyze the piecewise function [tex]\( f(x) \)[/tex] and verify the truth of each statement:
### Piecewise Function
[tex]\[ f(x)=\left\{\begin{array}{ll} -\frac{1}{4} x^2 + 6x + 36, & x < -2 \\ 4x - 15, & -2 \leq x < 4 \\ 3^{x-4}, & x > 4 \end{array}\right. \][/tex]
### Statement 1: The graph crosses the y-axis at [tex]\((0, -15)\)[/tex].
To find where the graph crosses the y-axis, we substitute [tex]\( x = 0 \)[/tex] into the corresponding interval for [tex]\( x \)[/tex].
For [tex]\( -2 \leq 0 < 4 \)[/tex]:
[tex]\[ f(0) = 4(0) - 15 = -15 \][/tex]
Thus, the graph crosses the y-axis at [tex]\((0, -15)\)[/tex], so this statement is true.
### Statement 2: The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex].
We need to check the value of [tex]\( f(x) \)[/tex] from the left and from the right at [tex]\( x = -2 \)[/tex] to see if they are equal.
From the left ([tex]\( x < -2 \)[/tex]):
[tex]\[ \lim_{x \to -2^-} f(x) = -\frac{1}{4} (-2)^2 + 6(-2) + 36 \][/tex]
[tex]\[ = -1 + (-12) + 36 = 23 \][/tex]
From the right ([tex]\( -2 \leq x < 4 \)[/tex]):
[tex]\[ \lim_{x \to -2^+} f(x) = 4(-2) - 15 \][/tex]
[tex]\[ = -8 - 15 = -23 \][/tex]
Since [tex]\( 23 \neq -23 \)[/tex], the function is discontinuous at [tex]\( x = -2 \)[/tex]. Therefore, this statement is true.
### Statement 3: The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
For [tex]\( x > 4 \)[/tex]:
[tex]\[ f(x) = 3^{x-4} \][/tex]
To determine if the graph is increasing, we need to check if the derivative is positive for [tex]\( x > 4 \)[/tex].
[tex]\[ \frac{d}{dx} f(x) = \frac{d}{dx} 3^{x-4} = 3^{x-4} \ln(3) \][/tex]
Since [tex]\( 3^{x-4} \)[/tex] and [tex]\( \ln(3) \)[/tex] are both positive for all [tex]\( x > 4 \)[/tex], the derivative is positive, and the function is indeed increasing over this interval. Hence, this statement is true.
### Statement 4: The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex].
For [tex]\(-\infty < x < -2\)[/tex]:
[tex]\[ f(x) = -\frac{1}{4} x^2 + 6x + 36 \][/tex]
The derivative gives us:
[tex]\[ \frac{d}{dx} f(x) = -\frac{1}{2} x + 6 \][/tex]
For [tex]\( -12 < x < -2 \)[/tex]:
[tex]\[ -\frac{1}{2} x + 6 \][/tex]
Substitute any [tex]\( x \)[/tex] in [tex]\( (-12, -2) \)[/tex] to verify if the derivative is negative.
Let's test with [tex]\( x = -3 \)[/tex]:
[tex]\[ -\frac{1}{2}(-3) + 6 = \frac{3}{2} + 6 = 7.5 \][/tex]
This value is positive rather than negative, indicating the function is actually increasing, not decreasing, in the interval [tex]\( (-12, -2) \)[/tex]. Therefore, this statement is false.
### Statement 5: The domain of the function is all real numbers.
The function [tex]\( f(x) \)[/tex] is defined for all [tex]\( x \in \mathbb{R} \)[/tex]. Each piece provided covers all possible [tex]\( x \)[/tex] values within their respective intervals, so the function is defined for all real numbers. Thus, this statement is true.
### Summary Table
[tex]\[ \begin{tabular}{|l|l|l|} \hline \text{The graph crosses the } y \text{-axis at } (0, -15) & \text{true} & \text{false} \\ \hline \text{The graph has a point of discontinuity at } x = -2 & \text{true} & \text{false} \\ \hline \text{The graph is increasing over the interval } (4, \infty) & \text{true} & \text{false} \\ \hline \text{The graph is decreasing over the interval } (-12, -2) & \text{true} & \text{false} \\ \hline \text{The domain of the function is all real numbers} & \text{true} & \text{false} \\ \hline \end{tabular} \][/tex]
### Piecewise Function
[tex]\[ f(x)=\left\{\begin{array}{ll} -\frac{1}{4} x^2 + 6x + 36, & x < -2 \\ 4x - 15, & -2 \leq x < 4 \\ 3^{x-4}, & x > 4 \end{array}\right. \][/tex]
### Statement 1: The graph crosses the y-axis at [tex]\((0, -15)\)[/tex].
To find where the graph crosses the y-axis, we substitute [tex]\( x = 0 \)[/tex] into the corresponding interval for [tex]\( x \)[/tex].
For [tex]\( -2 \leq 0 < 4 \)[/tex]:
[tex]\[ f(0) = 4(0) - 15 = -15 \][/tex]
Thus, the graph crosses the y-axis at [tex]\((0, -15)\)[/tex], so this statement is true.
### Statement 2: The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex].
We need to check the value of [tex]\( f(x) \)[/tex] from the left and from the right at [tex]\( x = -2 \)[/tex] to see if they are equal.
From the left ([tex]\( x < -2 \)[/tex]):
[tex]\[ \lim_{x \to -2^-} f(x) = -\frac{1}{4} (-2)^2 + 6(-2) + 36 \][/tex]
[tex]\[ = -1 + (-12) + 36 = 23 \][/tex]
From the right ([tex]\( -2 \leq x < 4 \)[/tex]):
[tex]\[ \lim_{x \to -2^+} f(x) = 4(-2) - 15 \][/tex]
[tex]\[ = -8 - 15 = -23 \][/tex]
Since [tex]\( 23 \neq -23 \)[/tex], the function is discontinuous at [tex]\( x = -2 \)[/tex]. Therefore, this statement is true.
### Statement 3: The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
For [tex]\( x > 4 \)[/tex]:
[tex]\[ f(x) = 3^{x-4} \][/tex]
To determine if the graph is increasing, we need to check if the derivative is positive for [tex]\( x > 4 \)[/tex].
[tex]\[ \frac{d}{dx} f(x) = \frac{d}{dx} 3^{x-4} = 3^{x-4} \ln(3) \][/tex]
Since [tex]\( 3^{x-4} \)[/tex] and [tex]\( \ln(3) \)[/tex] are both positive for all [tex]\( x > 4 \)[/tex], the derivative is positive, and the function is indeed increasing over this interval. Hence, this statement is true.
### Statement 4: The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex].
For [tex]\(-\infty < x < -2\)[/tex]:
[tex]\[ f(x) = -\frac{1}{4} x^2 + 6x + 36 \][/tex]
The derivative gives us:
[tex]\[ \frac{d}{dx} f(x) = -\frac{1}{2} x + 6 \][/tex]
For [tex]\( -12 < x < -2 \)[/tex]:
[tex]\[ -\frac{1}{2} x + 6 \][/tex]
Substitute any [tex]\( x \)[/tex] in [tex]\( (-12, -2) \)[/tex] to verify if the derivative is negative.
Let's test with [tex]\( x = -3 \)[/tex]:
[tex]\[ -\frac{1}{2}(-3) + 6 = \frac{3}{2} + 6 = 7.5 \][/tex]
This value is positive rather than negative, indicating the function is actually increasing, not decreasing, in the interval [tex]\( (-12, -2) \)[/tex]. Therefore, this statement is false.
### Statement 5: The domain of the function is all real numbers.
The function [tex]\( f(x) \)[/tex] is defined for all [tex]\( x \in \mathbb{R} \)[/tex]. Each piece provided covers all possible [tex]\( x \)[/tex] values within their respective intervals, so the function is defined for all real numbers. Thus, this statement is true.
### Summary Table
[tex]\[ \begin{tabular}{|l|l|l|} \hline \text{The graph crosses the } y \text{-axis at } (0, -15) & \text{true} & \text{false} \\ \hline \text{The graph has a point of discontinuity at } x = -2 & \text{true} & \text{false} \\ \hline \text{The graph is increasing over the interval } (4, \infty) & \text{true} & \text{false} \\ \hline \text{The graph is decreasing over the interval } (-12, -2) & \text{true} & \text{false} \\ \hline \text{The domain of the function is all real numbers} & \text{true} & \text{false} \\ \hline \end{tabular} \][/tex]
