Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Let's analyze each statement about the function [tex]\( f \)[/tex] given by:
[tex]\[ f(x)=\left\{ \begin{array}{ll} -\frac{1}{4} x^2+6 x+36, & x < -2 \\ 4 x-15, & -2 \leq x < 4 \\ 3^{x-4}, & x > 4 \\ \end{array} \right. \][/tex]
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
To determine if the graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex], we need to evaluate [tex]\( f(0) \)[/tex].
For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 4 \cdot 0 - 15 = -15 \][/tex]
Since [tex]\( f(0) = -15 \)[/tex], the graph indeed crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
2. The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex].
To check for a point of discontinuity at [tex]\( x = -2 \)[/tex], we need to evaluate the left-hand limit and the right-hand limit at [tex]\( x = -2 \)[/tex].
Left-hand limit as [tex]\( x \to -2^- \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} f(x) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = -1 - 12 + 36 = 23 \][/tex]
Right-hand limit as [tex]\( x \to -2^+ \)[/tex]:
[tex]\[ \lim_{{x \to -2^+}} f(x) = 4(-2) - 15 = -8 - 15 = -23 \][/tex]
Since the left-hand limit (23) is not equal to the right-hand limit (-23), there is a point of discontinuity at [tex]\( x = -2 \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
3. The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
For [tex]\( x > 4 \)[/tex], the function is given by [tex]\( f(x) = 3^{x-4} \)[/tex]. Exponential functions of the form [tex]\( 3^{x-4} \)[/tex] are strictly increasing.
Therefore, the graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
4. The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex].
On the interval [tex]\( (-12, -2) \)[/tex], the function is given by [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex].
To determine if the function is decreasing, we need to check the derivative of this part and ensure it's negative over [tex]\( (-12, -2) \)[/tex].
Evaluating the derivative:
[tex]\[ f'(x) = -\frac{1}{2} x + 6 \][/tex]
At [tex]\( x = -2 \)[/tex]:
[tex]\[ f'(-2) = -\frac{1}{2}(-2) + 6 = 1 + 6 = 7 \][/tex]
Since [tex]\( f'(x) > 0 \)[/tex] in the interval [tex]\( (-12, -2) \)[/tex], the function is increasing, not decreasing.
[tex]\[ \text{Answer: false} \][/tex]
5. The domain of the function is all real numbers.
The function [tex]\( f \)[/tex] is defined for all [tex]\( x \)[/tex] in the real numbers, as each piece of the function covers all real intervals without any restrictions or undefined values.
[tex]\[ \text{Answer: true} \][/tex]
Summarizing the answers:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{The graph crosses the } y \text{-axis at } (0,-15). & \text{true} & \text{false} \\ \hline \text{The graph has a point of discontinuity at } x =-2. & \text{true} & \text{false} \\ \hline \text{The graph is increasing over the interval } (4, \infty). & \text{true} & \text{false} \\ \hline \text{The graph is decreasing over the interval } (-12,-2). & \text{true} & \text{false} \\ \hline \text{The domain of the function is all real numbers}. & \text{true} & \text{false} \\ \hline \end{array} \][/tex]
[tex]\[ f(x)=\left\{ \begin{array}{ll} -\frac{1}{4} x^2+6 x+36, & x < -2 \\ 4 x-15, & -2 \leq x < 4 \\ 3^{x-4}, & x > 4 \\ \end{array} \right. \][/tex]
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
To determine if the graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex], we need to evaluate [tex]\( f(0) \)[/tex].
For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 4 \cdot 0 - 15 = -15 \][/tex]
Since [tex]\( f(0) = -15 \)[/tex], the graph indeed crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
2. The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex].
To check for a point of discontinuity at [tex]\( x = -2 \)[/tex], we need to evaluate the left-hand limit and the right-hand limit at [tex]\( x = -2 \)[/tex].
Left-hand limit as [tex]\( x \to -2^- \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} f(x) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = -1 - 12 + 36 = 23 \][/tex]
Right-hand limit as [tex]\( x \to -2^+ \)[/tex]:
[tex]\[ \lim_{{x \to -2^+}} f(x) = 4(-2) - 15 = -8 - 15 = -23 \][/tex]
Since the left-hand limit (23) is not equal to the right-hand limit (-23), there is a point of discontinuity at [tex]\( x = -2 \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
3. The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
For [tex]\( x > 4 \)[/tex], the function is given by [tex]\( f(x) = 3^{x-4} \)[/tex]. Exponential functions of the form [tex]\( 3^{x-4} \)[/tex] are strictly increasing.
Therefore, the graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
4. The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex].
On the interval [tex]\( (-12, -2) \)[/tex], the function is given by [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex].
To determine if the function is decreasing, we need to check the derivative of this part and ensure it's negative over [tex]\( (-12, -2) \)[/tex].
Evaluating the derivative:
[tex]\[ f'(x) = -\frac{1}{2} x + 6 \][/tex]
At [tex]\( x = -2 \)[/tex]:
[tex]\[ f'(-2) = -\frac{1}{2}(-2) + 6 = 1 + 6 = 7 \][/tex]
Since [tex]\( f'(x) > 0 \)[/tex] in the interval [tex]\( (-12, -2) \)[/tex], the function is increasing, not decreasing.
[tex]\[ \text{Answer: false} \][/tex]
5. The domain of the function is all real numbers.
The function [tex]\( f \)[/tex] is defined for all [tex]\( x \)[/tex] in the real numbers, as each piece of the function covers all real intervals without any restrictions or undefined values.
[tex]\[ \text{Answer: true} \][/tex]
Summarizing the answers:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{The graph crosses the } y \text{-axis at } (0,-15). & \text{true} & \text{false} \\ \hline \text{The graph has a point of discontinuity at } x =-2. & \text{true} & \text{false} \\ \hline \text{The graph is increasing over the interval } (4, \infty). & \text{true} & \text{false} \\ \hline \text{The graph is decreasing over the interval } (-12,-2). & \text{true} & \text{false} \\ \hline \text{The domain of the function is all real numbers}. & \text{true} & \text{false} \\ \hline \end{array} \][/tex]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.