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Inverse Functions Simplified Revision Notes

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2.8.3 Inverse Functions

Inverse of a Function

A function only has an inverse if it is one-to-one (11-to-11).

  • Example: f(x)=x2,xRf(x) = x^2, x \in \mathbb{R} has no inverse.
  • Function is many-to-one, therefore no inverse exists. image

However, we can "artificially" restrict the domain of a function to make it one-to-one (11-to-11), thus forcing the existence of an inverse.

  • Example: f(x)=x2,x0f(x) = x^2, x \geq 0
  • One-to-one, therefore has an inverse.
image

The inverse of a function is denoted f1(x)f^{-1}(x) for a function f(x)f(x).

  • For f(x)=x2,x0f(x) = x^2, x \geq 0: f1(x)=xf^{-1}(x) = \sqrt{x}

Steps to Find the Inverse of a Function

infoNote

Example: Find the inverse function of f(x)=ex,xRf(x) = e^x, x \in \mathbb{R}.

  1. Write as y=f(x)y = f(x) and swap xx's and yy's.
y=exy = e^xx=eyx = e^y

2) Rearrange to obtain yy:

x=ey    lnx=ln(ey)x = e^y \implies \ln x = \ln (e^y)    lnx=y\implies \ln x = y

3) Rewrite f1(x)f^{-1}(x):

f1(x)=ln(x)f^{-1}(x) = \ln(x)

IMPORTANT: You should always write the domain of an inverse. This is the range of the original function.

4) Write the domain:

Range of f(x):f(x)=exf(x): f(x) = e^x

Range: f(x)>0f(x) > 0

Range in terms of f(x)f(x).

Domain of f1(x) is x>0(Domain in terms of x)\therefore \text{Domain of } f^{-1}(x) \text{ is } x > 0 \quad (\text{Domain in terms of } x)

infoNote

Example: Find the inverse of f(x)=2x+3,x>3f(x) = 2x + 3, x > 3

y=2x+3y = 2x + 3x=2y+3    x3=2y    y=x32x = 2y + 3 \implies x - 3 = 2y \implies y = \frac{x - 3}{2}f1(x)=x32\therefore f^{-1}(x) = \frac{x-3}{2}

Let x=3x = 3, then:

f(3)=2(3)+3=9f(3) = 2(3) + 3 = 9Range in f(x)9\therefore \text{Range in } f(x) \ge 9f1(x)=x32,x9\therefore f^{-1}(x) = \frac{x-3}{2}, \quad x \ge 9

Geometric Relationship Between a Function and its Inverse

To find the inverse of a function, we simply swap its xx and y values. This is the equivalent of reflecting the graph through the line y=xy = x.

image

Graphical Representation:

  • The graph shows the functions y=exy = e^x (red curve) and y=ln(x)y = \ln(x) (blue curve) with the line y=xy = x as a dotted diagonal line.
  • Key points marked on the graph:
    • (0,1) on y=exy = e^x
    • (1,0) on y=ln(x)y = \ln(x)
infoNote

Advice: 2. Turn the page around so that you are looking directly up the mirror line y=xy = x. 3. Swap the coordinates of any of the points of intersections.


infoNote

Example: Given that f(x)=6x4,xRf(x) = 6x - 4, x \in \mathbb{R}, find where y=f(x)y = f(x) intersects y=f1(x)y = f^{-1}(x). Since y=f(x)y = f(x) and y=f1(x)y = f^{-1}(x) are reflections of each other in the line y=xy = x, if they intersect, they do so on the line y=xy = x.

Therefore, solving y=xy = x and y=6x4y = 6x - 4 simultaneously:

x=6x45x=4x=45y=45\begin{align*} x &= 6x - 4 \\ 5x &= 4 \\ x &= \frac{4}{5} \\ y &= \frac{4}{5} \end{align*}

Thus, the point of intersection is (45,45)\left( \dfrac{4}{5}, \dfrac{4}{5} \right).


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