Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Sure! Let's go through this problem step-by-step.
### (a) Find the inverse of [tex]\( f(x) = x^2 + 10 \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
1. We start with the function:
[tex]\[ y = x^2 + 10 \][/tex]
2. We need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]. First, isolate [tex]\( x^2 \)[/tex]:
[tex]\[ y - 10 = x^2 \][/tex]
3. Next, solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x = \sqrt{y - 10} \][/tex]
Since [tex]\( x \geq 0 \)[/tex], we take the non-negative root. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x - 10} \][/tex]
### (b) Find the domain and the range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
#### Domain and range of [tex]\( f(x) = x^2 + 10 \)[/tex]
1. Domain of [tex]\( f \)[/tex]:
Since [tex]\( x \geq 0 \)[/tex], the domain of [tex]\( f \)[/tex] is:
[tex]\[ 0 \leq x < \infty \][/tex]
In interval notation, this is:
[tex]\[ \text{Domain of } f: (0, \infty) \][/tex]
2. Range of [tex]\( f \)[/tex]:
The minimum value of [tex]\( x^2 \)[/tex] is 0 (since [tex]\( x \geq 0 \)[/tex]), and hence the minimum value of [tex]\( f(x) \)[/tex] is:
[tex]\[ 10 \leq f(x) < \infty \][/tex]
In interval notation, this is:
[tex]\[ \text{Range of } f: (10, \infty) \][/tex]
#### Domain and range of [tex]\( f^{-1}(x) = \sqrt{x - 10} \)[/tex]
1. Domain of [tex]\( f^{-1} \)[/tex]:
Since the argument [tex]\( x - 10 \)[/tex] must be non-negative for the square root to be defined, we have:
[tex]\[ x \geq 10 \][/tex]
In interval notation, this is:
[tex]\[ \text{Domain of } f^{-1}: (10, \infty) \][/tex]
2. Range of [tex]\( f^{-1} \)[/tex]:
The output of the square root function [tex]\( \sqrt{x - 10} \)[/tex], given [tex]\( x - 10 \geq 0 \)[/tex], is:
[tex]\[ 0 \leq \sqrt{x - 10} < \infty \][/tex]
In interval notation, this is:
[tex]\[ \text{Range of } f^{-1}: (0, \infty) \][/tex]
### (c) Graph [tex]\( f, f^{-1} \)[/tex], and [tex]\( y = x \)[/tex] on the same coordinate axes
To visualize:
- The graph of [tex]\( f(x) = x^2 + 10 \)[/tex] is a parabola shifted up by 10 units.
- The graph of [tex]\( f^{-1}(x) = \sqrt{x - 10} \)[/tex] is a half-parabola starting from [tex]\( x = 10 \)[/tex].
- The graph of [tex]\( y = x \)[/tex] is a straight line with a slope of 1 passing through the origin.
These details would typically be plotted as follows:
1. [tex]\( f(x) = x^2 + 10 \)[/tex]: A parabola starting at the point (0, 10) and opening upwards.
2. [tex]\( f^{-1}(x) = \sqrt{x - 10} \)[/tex]: A curve starting at (10, 0) and moving to the right and upwards.
3. [tex]\( y = x \)[/tex]: A diagonal line passing through the origin.
By plotting these on the same coordinate axes, you can see how the functions are reflections of each other across the line [tex]\( y = x \)[/tex].
### Summary:
(a)
[tex]\[ f^{-1}(x) = \sqrt{x - 10} \][/tex]
(b)
Domain of [tex]\( f \)[/tex]: [tex]\( (0, \infty) \)[/tex]
Range of [tex]\( f \)[/tex]: [tex]\( (10, \infty) \)[/tex]
Domain of [tex]\( f^{-1} \)[/tex]: [tex]\( (10, \infty) \)[/tex]
Range of [tex]\( f^{-1} \)[/tex]: [tex]\( (0, \infty) \)[/tex]
(c) Graphically plot [tex]\( f(x) \)[/tex], [tex]\( f^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex] on the same set of axes.
### (a) Find the inverse of [tex]\( f(x) = x^2 + 10 \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
1. We start with the function:
[tex]\[ y = x^2 + 10 \][/tex]
2. We need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]. First, isolate [tex]\( x^2 \)[/tex]:
[tex]\[ y - 10 = x^2 \][/tex]
3. Next, solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x = \sqrt{y - 10} \][/tex]
Since [tex]\( x \geq 0 \)[/tex], we take the non-negative root. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x - 10} \][/tex]
### (b) Find the domain and the range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
#### Domain and range of [tex]\( f(x) = x^2 + 10 \)[/tex]
1. Domain of [tex]\( f \)[/tex]:
Since [tex]\( x \geq 0 \)[/tex], the domain of [tex]\( f \)[/tex] is:
[tex]\[ 0 \leq x < \infty \][/tex]
In interval notation, this is:
[tex]\[ \text{Domain of } f: (0, \infty) \][/tex]
2. Range of [tex]\( f \)[/tex]:
The minimum value of [tex]\( x^2 \)[/tex] is 0 (since [tex]\( x \geq 0 \)[/tex]), and hence the minimum value of [tex]\( f(x) \)[/tex] is:
[tex]\[ 10 \leq f(x) < \infty \][/tex]
In interval notation, this is:
[tex]\[ \text{Range of } f: (10, \infty) \][/tex]
#### Domain and range of [tex]\( f^{-1}(x) = \sqrt{x - 10} \)[/tex]
1. Domain of [tex]\( f^{-1} \)[/tex]:
Since the argument [tex]\( x - 10 \)[/tex] must be non-negative for the square root to be defined, we have:
[tex]\[ x \geq 10 \][/tex]
In interval notation, this is:
[tex]\[ \text{Domain of } f^{-1}: (10, \infty) \][/tex]
2. Range of [tex]\( f^{-1} \)[/tex]:
The output of the square root function [tex]\( \sqrt{x - 10} \)[/tex], given [tex]\( x - 10 \geq 0 \)[/tex], is:
[tex]\[ 0 \leq \sqrt{x - 10} < \infty \][/tex]
In interval notation, this is:
[tex]\[ \text{Range of } f^{-1}: (0, \infty) \][/tex]
### (c) Graph [tex]\( f, f^{-1} \)[/tex], and [tex]\( y = x \)[/tex] on the same coordinate axes
To visualize:
- The graph of [tex]\( f(x) = x^2 + 10 \)[/tex] is a parabola shifted up by 10 units.
- The graph of [tex]\( f^{-1}(x) = \sqrt{x - 10} \)[/tex] is a half-parabola starting from [tex]\( x = 10 \)[/tex].
- The graph of [tex]\( y = x \)[/tex] is a straight line with a slope of 1 passing through the origin.
These details would typically be plotted as follows:
1. [tex]\( f(x) = x^2 + 10 \)[/tex]: A parabola starting at the point (0, 10) and opening upwards.
2. [tex]\( f^{-1}(x) = \sqrt{x - 10} \)[/tex]: A curve starting at (10, 0) and moving to the right and upwards.
3. [tex]\( y = x \)[/tex]: A diagonal line passing through the origin.
By plotting these on the same coordinate axes, you can see how the functions are reflections of each other across the line [tex]\( y = x \)[/tex].
### Summary:
(a)
[tex]\[ f^{-1}(x) = \sqrt{x - 10} \][/tex]
(b)
Domain of [tex]\( f \)[/tex]: [tex]\( (0, \infty) \)[/tex]
Range of [tex]\( f \)[/tex]: [tex]\( (10, \infty) \)[/tex]
Domain of [tex]\( f^{-1} \)[/tex]: [tex]\( (10, \infty) \)[/tex]
Range of [tex]\( f^{-1} \)[/tex]: [tex]\( (0, \infty) \)[/tex]
(c) Graphically plot [tex]\( f(x) \)[/tex], [tex]\( f^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex] on the same set of axes.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
