Answer :
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.
