Find the Inverse Function from Graphs

Graphical Explanations and Step-by-Step Solved Questions

Finding the values of an inverse function from a graph is a visual process. It relies on the fundamental property that if an original function maps $x$ to $y$, the inverse function maps $y$ back to $x$.

The Inverse Relationship

If $f$ is a function and $f^{-1}$ is its inverse, their relationship is defined as:

\[ f(a) = b \iff a = f^{-1}(b) \]

Visual Strategy: To find $f^{-1}(b)$, you are essentially looking for the $x$-value that produces a $y$-value of $b$.

Worked Example: Curved Relation $f(x)$

Use the graph of function $f$ below to find:

Graph of function f
View Solutions

a) $f^{-1}(5)$: Locate $y = 5$ on the y-axis. Move horizontally to the graph and then down to the x-axis. We find $x = 3$. Therefore, $f^{-1}(5) = 3$.

Visual solution for f^-1(5)

b) $f^{-1}(0)$: At $y = 0$, the graph is at $x = 2$. Thus, $f^{-1}(0) = 2$.

c) $f^{-1}(-3)$: At $y = -3$, the graph is at $x = 1$. Thus, $f^{-1}(-3) = 1$.

d) $f^{-1}(-4)$: At $y = -4$, the graph is at $x = 0$. Thus, $f^{-1}(-4) = 0$.

e) $f^{-1}(-5)$: The graph does not reach $y = -5$. This value is undefined.

Practice Questions

Question 1: Linear Relation $g(x)$

Find: $g^{-1}(6)$, $g^{-1}(0)$, $g^{-1}(-2)$, and $g^{-1}(8)$.

Graph of function g
View Solution

a) $g^{-1}(6)$: The $x$-value for $y = 6$ is 2. So, $g^{-1}(6) = 2$.

Solution for g^-1

b) $g^{-1}(0)$: The graph crosses the x-axis ($y=0$) at $x = -1$. So, $g^{-1}(0) = -1$.

c) $g^{-1}(-2)$: The graph is at $x = -2$ when $y = -2$. So, $g^{-1}(-2) = -2$.

d) $g^{-1}(8)$: The graph's maximum height is $y = 6$. Therefore, $g^{-1}(8)$ is undefined.

Question 2: Trigonometric Relation $h(x)$

Find: $h^{-1}(1)$, $h^{-1}(0)$, and $h^{-1}(-1)$.

Graph of trigonometric function h
View Solution

a) $h^{-1}(1)$: The graph is at height 1 when $x = 0$. So, $h^{-1}(1) = 0$.

b) $h^{-1}(0)$: The graph crosses $y=0$ at $x = \pi/2$. So, $h^{-1}(0) = \pi/2$.

c) $h^{-1}(-1)$: The graph reaches $-1$ at $x = \pi$. So, $h^{-1}(-1) = \pi$.

d) $h^{-1}(2)$: The maximum $y$-value is 1. Therefore, $h^{-1}(2)$ is undefined.

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