In earlier grades, you studied various types of functions and understood the effect of various parameters in the general equation. In this section, we will consider inverse functions.
An inverse function is a function which "does the reverse" of a given function. More formally, if ff is a function with domain XX, then f-1f-1 is its inverse function if and only if for every x∈Xx∈X we have:
f
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1
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f
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=
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f
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f
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x
(5)A simple way to think about this is that a function, say y=f(x)y=f(x), gives you a yy-value if you substitute an xx-value into f(x)f(x). The inverse function tells you tells you which xx-value was used to get a particular yy-value when you substitue the yy-value into f-1(x)f-1(x). There are some things which can complicate this for example, think about a sinsin function, there are many xx-values that give you a peak as the function oscillates. This means that the inverse of a sinsin function would be tricky to define because if you substitute the peak yy-value into it you won't know which of the xx-values was used to get the peak.
y
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f
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we have a function
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1
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we substitute a specific x-value into the function to get a specific y-value
consider the inverse function
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x
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substituting the specific y-value into the inverse should return the specific x-value
=
f
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y
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f
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we have a function
y
1
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f
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x
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we substitute a specific x-value into the function to get a specific y-value
consider the inverse function
x
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f
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1
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y
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x
=
f
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1
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substituting the specific y-value into the inverse should return the specific x-value
=
f
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y
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=
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1
(6)This works both ways, if we don't have any complications like in the case of the sinsin function, so we can write:
f
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f
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f
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(7)For example, if the function x→3x+2x→3x+2 is given, then its inverse function is x→(x-2)3x→(x-2)3. This is usually written as:
f
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f
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→
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3
f
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3
(8)The superscript "-1" is not an exponent.
If a function ff has an inverse then ff is said to be invertible.
If ff is a real-valued function, then for ff to have a valid inverse, it must pass the horizontal line test, that is a horizontal line y=ky=k placed anywhere on the graph of ff must pass through ff exactly once for all real kk.
It is possible to work around this condition, by defining a “multi-valued“ function as an inverse.
If one represents the function ff graphically in a xyxy-coordinate system, the inverse function of the equation of a straight line, f-1f-1, is the reflection of the graph of ff across the line y=xy=x.
Algebraically, one computes the inverse function of ff by solving the equation
y
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f
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y
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(9)for xx, and then exchanging yy and xx to get
y
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y
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(10)The inverse function of y=ax+qy=ax+q is determined by solving for xx as:
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a
(11)Therefore the inverse of y=ax+qy=ax+q is y=1ax-qay=1ax-qa.
The inverse function of a straight line is also a straight line, except for the case where the straight line is a perfectly horizontal line, in which case the inverse is undefined.
For example, the straight line equation given by y=2x-3y=2x-3 has as inverse the function, y=12x+32y=12x+32. The graphs of these functions are shown in Figure 3. It can be seen that the two graphs are reflections of each other across the line y=xy=x.
We have seen that the domain of a function of the form y=ax+qy=ax+q is {x:x∈R}{x:x∈R} and the range is {y:y∈R}{y:y∈R}. Since the inverse function of a straight line is also a straight line, the inverse function will have the same domain and range as the original function.
The general form of the inverse function of the form y=ax+qy=ax+q is y=1ax-qay=1ax-qa.
By setting x=0x=0 we have that the yy-intercept is yint=-qayint=-qa. Similarly, by setting y=0y=0 we have that the xx-intercept is xint=qxint=q.
It is interesting to note that if f(x)=ax+qf(x)=ax+q, then f-1(x)=1ax-qaf-1(x)=1ax-qa and the yy-intercept of f(x)f(x) is the xx-intercept of f-1(x)f-1(x) and the xx-intercept of f(x)f(x) is the yy-intercept of f-1(x)f-1(x).
- Given f(x)=2x-3f(x)=2x-3, find f-1(x)f-1(x)
- Consider the function f(x)=3x-7f(x)=3x-7.
- Is the relation a function?
- If it is a function, identify the domain and range.
- Sketch the graph of the function f(x)=3x-1f(x)=3x-1 and its inverse on the same set of axes.
- The inverse of a function is f-1(x)=2x-4f-1(x)=2x-4, what is the function f(x)f(x)?
The inverse relation, possibly a function, of y=ax2y=ax2 is determined by solving for xx as:
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2
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(12)We see that the inverse ”function” of y=ax2y=ax2 is not a function because it fails the vertical line test. If we draw a vertical line through the graph of f-1(x)=±xf-1(x)=±x, the line intersects the graph more than once. There has to be a restriction on the domain of a parabola for the inverse to also be a function. Consider the function f(x)=-x2+9f(x)=-x2+9. The inverse of ff can be found by witing f(y)=xf(y)=x. Then
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(13)If we restrict the domain of f(x)f(x) to be x≥0x≥0, then 9-x9-x is a function. If the restriction on the domain of ff is x≤0x≤0 then -9-x-9-x would be a function, inverse to ff.
- The graph of f-1f-1 is shown. Find the equation of ff, given that the graph of ff is a parabola. (Do not simplify your answer)
- f(x)=2x2f(x)=2x2.
- Draw the graph of ff and state its domain and range.
- Find f-1f-1 and, if it exists, state the domain and range.
- What must the domain of ff be, so that f-1f-1 is a function ?
- Sketch the graph of x=-10-y2x=-10-y2. Label a point on the graph other than the intercepts with the axes.
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- Sketch the graph of y=x2y=x2 labelling a point other than the origin on your graph.
- Find the equation of the inverse of the above graph in the form y=...y=....
- Now sketch the graph of y=xy=x.
- The tangent to the graph of y=xy=x at the point A(9;3) intersects the xx-axis at B. Find the equation of this tangent and hence or otherwise prove that the yy-axis bisects the straight line AB.
- Given: g(x)=-1+xg(x)=-1+x, find the inverse of g(x)g(x) in the form g-1(x)=...g-1(x)=....
The inverse function of y=ax2y=ax2 is determined by solving for xx as follows:
y
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x
log
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log
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log
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∴
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y
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log
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∴
x
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(14)The inverse of y=10xy=10x is x=10yx=10y, which we write as y=log(x)y=log(x). Therefore, if f(x)=10xf(x)=10x, then f-1=log(x)f-1=log(x).
The exponential function and the logarithmic function are inverses of each other; the graph of the one is the graph of the other, reflected in the line y=xy=x.
The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse.
- Given that f(x)=(15)xf(x)=(15)x, sketch the graphs of ff and f-1f-1 on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is ff and which is f-1f-1.
- Given that f(x)=4-xf(x)=4-x,
- Sketch the graphs of ff and f-1f-1 on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is ff and which is f-1f-1.
- Write f-1f-1 in the form y=...y=....
- Given g(x)=-1+xg(x)=-1+x, find the inverse of g(x)g(x) in the form g-1(x)=...g-1(x)=...
- Answer the following questions:
- Sketch the graph of y=x2y=x2, labeling a point other than the origin on your graph.
- Find the equation of the inverse of the above graph in the form y=...y=...
- Now, sketch y=xy=x.
- The tangent to the graph of y=xy=x at the point A(9;3)A(9;3) intersects the xx-axis at BB. Find the equation of this tangent, and hence, or otherwise, prove that the yy-axis bisects the straight line ABAB.
