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Here are a few recommended readings before getting started with this lesson.
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.
The corresponding inverse relation is not a function because it has multiple outputs assigned to the same input, which goes against the definition of function.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
\CommutativePropAdd
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
\CommutativePropAdd
Subtract term
a2±2ab+b2=(a±b)2
LHS−90=RHS−90
LHS⋅(-160)=RHS⋅(-160)
Distribute -160
\CommutativePropAdd
LHS=RHS
LHS+80=RHS+80
Rearrange equation