How does the domain and range of a function relate to its inverse?
A
a
Example Solution: D: x≥ -2
B
b
Example Solution: y^(- 1)=-2+sqrt(x+7/3)
C
c
Example Solution: D: x≥ -7 and R: y≥ -2
Practice makes perfect
a
The function is given in its graphing form. Since the question is about the graph, let's start by graphing the function in its entirety. For a function in graphing form y=a(x-h)^2+k, also known as the vertex form, the vertex is (h,k) and the sign of a gives the direction.
y=3(x+2)^2-7Here, that means that the vertex is (-2,-7), and since 3 is positive, the parabola opens upward.
One way of dividing this graph in half is along its axis of symmetry. It has the same x-value as the vertex, x=-2. Let's choose the right half.
The x-values to the right of the axis of symmetry are x greater than -2, so the domain for the right part of the graph is x≥ -2.
b
Let start by finding the inverse. That means switching x and y.
Are we finished now? No. Remember that we chose the domain x≥-2 for the function. When finding an inverse function the domain and range are switched. That means that the range for the inverse function must be y≥ -2. Thus, we can exclude the negative radical expression.
y^(- 1)=-2+sqrt(x+7/3)
c
In the previous exercise we established that the range is y≥-2. Let's look at the function rule for the inverse again.
y^(- 1)=-2+sqrt(x+7/3)
The radical expression is only defined if x+73 is non-negative, which means that x needs to be greater than or equal to -7. This means that the domain of the inverse is x≥ -7.
Note that the domain and range of the inverse are reversed compared to the original function.
The x-values for the function are greater than -2, and the y-values of y^(-1) are also greater than -2. Correspondingly, as the possible y-values the original function are greater than -7, the same is true for the possible x-values of the inverse.
