The domain is the set of all x-values for which the function is defined. The range is the set of all y-values a function gives.
B
b
When a function is reflected in a line of symmetry it is mapped onto itself. Can you draw such a line?
C
c
What is y at an x-intercept? What is x at a y-intercept?
D
d
For inverse functions, the domain is the range and range is the domain of the given function.
A
a
Domain: - ∞ < x < ∞
Range: - 3 < y < ∞
B
b
There is no line of symmetry.
C
c
x-intercept: (≈ 1.585,0) y-intercept: (0,- 2)
D
d
Graph:
Domain: - 3 < x < ∞ Range: - ∞ < x < ∞ x-intercept: (-2,0) y-intercept: (0,≈ 1.585)
Practice makes perfect
a
To be able to sketch the graph, we first need to find some points on it. When calculating the y-values we will use an alternative version of the equation.
y+3=2^x ⇔ y=2^x-3Let's make a table of values.
x
2^x-3
y
- 4
2^(- 4)-3
- 2.9375
- 1
2^(- 1)-3
- 2.5
0
2^0-3
- 2
1
2^1-3
-1
2
2^2-3
1
3
2^3-3
5
Next we will make a diagram and mark these points.
The graph can now be sketched by connecting these dots with a smooth curve.
The domain is the set of all x-values for which the function is defined. This function is defined for all x. The range is the set of all y-values a function gives. Since 2^x is greater than 0 for all x, the function's range is all y greater than - 3.
Domain: & - ∞ < x < ∞
Range: & - 3 < y < ∞
b
When a function is reflected in a line of symmetry it is mapped onto itself. It is not possible to draw such a line. Therefore, there is no line of symmetry.
c
At a y-intercept we know that x=0. We find it by substituting 0 for x. When doing this we will use y=2^x-3, which is an alternative version of the equation.
y=2^x-3 ⇒ y=2^0-3=- 2
Thus, the y-intercept is at (0,- 2). At an x-intercept we know that y=0. Since substituting 0 for y would lead to an equation we cannot solve at the moment, we will have to take a different approach. In the graph of y+3=2^y let's locate the x-intercept.
We will now find the x-intercept by guessing an x-value, which we will test in the function. We will then use the result to refine the guess. Using the graph, we make the first guess as x=1.5.
y=2^x-3 ⇓ y=2^(1.5)-3=- 0.17157...
From this we know that the x-intercept is slightly larger than our first guess. Let's try a few more x-values.
x
2^x-3
y
1.52
2^(1.52)-3
- 0.13208...
1.54
2^(1.54)-3
- 0.09205 ...
1.56
2^(1.56)-3
- 0.05146 ...
1.58
2^(1.58)-3
- 0.01030 ...
1.60
2^(1.60)-3
0.03043 ...
The x-intercept is somewhere between x=1.58 and x=1.60. Let's try x=1.59.
y=2^x-3 ⇓ y=2^(1.59)-3= 0.01049...
We can see that the x-intercept is somewhere in the interval 1.58
d
We want to sketch the graphs of the given function and its inverse. To do so, let's first recall the table of values we created in Part A.
x
y
- 4
- 2.9375
- 1
- 2.5
0
- 2
1
-1
2
1
3
5
Since the inverse of a function swaps its x- and y-values, we can easily use the table above to make a table of values for the inverse.
x
y
- 2.9375
- 4
- 2.5
- 1
- 2
0
-1
1
1
2
5
3
Let's mark these point on the same graph we made in Part A and connect them with a smooth curve.
To find the domain and range of the inverse function, we should recall the domain and range of the given function.
Domain: & - ∞ < x < ∞
Range: & - 3 < y < ∞
To get the sets for the inverse, we need to swap the domain and range.
Domain of the inverse: & - 3 < y < ∞
Range of the inverse: & - ∞ < x < ∞
Similarly, the x- and y-intercepts of the function become the y- and x-intercepts of the inverse. Let's recall those points for the given function that we found in Part C.
x-intercept: & (≈ 1.585,0)
y-intercept: & (0,- 2)
Now, let's swap them to find the intercepts of the inverse. Remember that we also need to swap the coordinates of each point, because x-intercepts must have a y-coordinate of 0 and y-intercepts must have an x-coordinate of 0.
x-intercept of the inverse: & (-2,0)
y-intercept of the inverse: & (0,≈ 1.585)
