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| 11 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Every function has an inverse relation. If this inverse relation is also a function, then it is called an inverse function. In other words, the inverse of a function f is another function f-1 such that they undo each other.
f(f-1(x))=xandf-1(f(x))=x
Also, if x is the input of a function f and y its corresponding output, then y is the input of f-1 and x its corresponding output.
f(x)=y⇔f-1(y)=x
Definition of First Function | Substitute Second Function | Simplify | |
---|---|---|---|
f(f-1(x))=?x | 2f-1(x)−3=?x | 2(2x+3)−3=?x | x=x ✓ |
f-1(f(x))=?x | 2f(x)+3=?x | 22x−3+3=?x | x=x ✓ |
Therefore, f and f-1 undo each other. The graphs of these functions are each other's reflection across the line y=x. This means that the points on the graph of f-1 are the reversed points on the graph of f.
There is a condition that a function must satisfy in order to be invertible.
This condition implies that if a function fails the Horizontal Line Test, then it is not invertible. However, some functions do not satisfy this condition for invertibility but can become invertible by restricting their domain. For example, consider the quadratic parent function f(x)=x2.
This function has different x-values where the same y-value is assigned. The foll how the function fails the Horizontal Line Test.
Now, the graph of f(x) will be reflected across the line y=x to visualize its inverse relation.
The inverse is not a function because it has input values with two different outputs assigned. The following graph further confirms this relation by showing how it fails the Vertical Line Test.
Therefore, f(x)=x2 is not invertible because its inverse is not a function. Now, consider if the domain of f(x) is restricted to x≥0. In that case, every x-value gets assigned a unique y-value. With the domain of f(x) defined this way, f-1(x)=x is a valid inverse function.
Consider a function f(x) that assigns the same output to multiple inputs.
Heichi is so fascinated about the invertibility of quadratic functions, that he started a study group.
He brought the following function rule.
The following applet shows different quadratic functions. The representation of these functions alternates between showing their graph and stating their function rule. Determine whether or not the given function is invertible in the indicated domain.
LHS⋅3=RHS⋅3
LHS+1=RHS+1
LHS/2=RHS/2
Rearrange equation
LHS−k=RHS−k
LHS/a=RHS/a
LHS=RHS
LHS+h=RHS+h
Rearrange equation
Rewrite y as f-1(x)
Domain: 0≤x≤3
Range: 0≤y≤3
However, all the x-values in the interval 0≤x≤3 have unique y-values associated. This is why restricting the function's domain to the interval 0≤x≤3 makes the function invertible.
LHS−45=RHS−45
LHS/(-5)=RHS/(-5)
Put minus sign in numerator
Distribute -1
Commutative Property of Addition
LHS=RHS
Rearrange equation
x=30
Subtract term
Calculate quotient
Calculate root
Round to 1 decimal place(s)
x=20
Subtract term
Calculate quotient
Calculate root
Round to 1 decimal place(s)
The following applet shows a quadratic function and a specific point on its graph. The domain of the given function is restricted and the function is invertible. Find the coordinates of the corresponding point on the graph of its inverse function. Write the answer rounding to two decimal places.
a=a+6400−6400
Commutative Property of Addition
Subtract term
a2±2ab+b2=(a±b)2
LHS−90=RHS−90
LHS⋅(-160)=RHS⋅(-160)
Distribute -160
Commutative Property of Addition
LHS=RHS
LHS+80=RHS+80
Rearrange equation
Domain of f(x) | Inverse Function |
---|---|
0≤x<∞ | f-1(x)=x |
-∞<x≤0 | f-1(x)=-x |
A great method to explore this is by using the following interactive graph. The graph gives various choices of domains for f(x)=x2. The function's corresponding inverse relation is shown by reflecting f(x) across y=x.
Describe the error in finding and graphing the inverse of the function f(x)=x+4.
Let's take a look at the given exercise, paying close attention to each step, to see if we can find the error.
As can be seen in the steps the student wrote, the student began by attempting to find the inverse of y=sqrt(x+4) and its graph. Let's mimic this step by also solving for the inverse of the function. To do so, we will start by switching x and y. y=sqrt(x+4) ⇒ x=sqrt(y+4) So far, the notes are correct. Let's continue checking the steps and solve for y.
Because the range of f is y≥ 0, the domain of the inverse must be restricted to x≥0 as well. This is the mistake — the student forgot to restrict the domain of the inverse. g(x)=x^2-4, where x≥0 We will graph the inverse in a correct way.
Consider the given equation for the height h above ground (in feet) of a kite with respect to the time since the kite starts to fall. h=-4t^2+60 We are asked to solve this equation for t. Let's do it!
Since a negative time does not make sense in the context of the problem, we will consider the positive t as the solution. t=sqrt(60-h)/2
We are also asked to find at what time the kite would be 40 feet above ground. We can do this by substituting this value for h into our formula and simplifying.
We can use a calculator to find this value rounded to one decimal place.
Therefore, the kite will be at 40 feet above ground about 2.2 seconds after it starts to fall.
Let's start by drawing the graph of the given function. To do so, we will first make a table of values.
x | x^2+3 | f(x)=x^2+3 |
---|---|---|
- 2 | ( - 2)^2+3 | 7 |
- 1 | ( - 1)^2+3 | 4 |
0 | 0^2+3 | 3 |
1 | 1^2+3 | 4 |
2 | 2^2+3 | 7 |
Let's now plot and connect the obtained points. Keep in mind that the given function is quadratic and, therefore, its graph is a parabola.
Let's now apply the Horizontal Line Test.
Since the graph does not pass the Horizontal Line Test, the inverse of f is not a function. However, if we restrict the domain of f to be all non-negative real numbers, its graph will pass the Horizontal Line Test.
Great! Since the obtained graph passes the Horizontal Line Test, we can conclude that restricting the domain of f to be all non-negative real numbers makes its inverse a function. Please note that there are more ways to restrict the domain to make f invertible.