Category: Math

Inverse Functions A function $latex f(x) = x + 3 $ from $latex A = \{1,\, 2,\, 3,\, 4\} $ to $latex B = \{4,\, 5,\, 6,\, 7\} $ can be written as: $latex f : \{(1,\, 4),\, (2,\, 5),\, (3,\, 6),\, (4,\, 7)\} $. The inverse function $latex f^{-1} $ is formed by interchanging the coordinates: $latex f^{-1} : \{(4,\, 1),\, (5,\, 2),\, (6,\, 3),\, (7,\, 4)\} $. The domain of $latex f $ equals the range of $latex f^{-1} $ and vice versa. $latex f(f^{-1}(x)) = x $ and $latex f^{-1}(f(x)) = x $. Inverse Function Definition $latex f(g(x)) = x $ and $latex g(f(x)) = x $. The inverse of $latex f $ is denoted $latex f^{-1} $. Key Observations If $latex g $ is the inverse of $latex f $, then $latex f $ is the inverse of $latex g $. The domain of $latex f^{-1} $ equals the range of $latex f $, and the range of $latex f^{-1} $ equals the domain of $latex f $. Not all functions have inverses, but when they do, it's unique. Subtraction undoes addition: $latex f(x) = x + c $, $latex f^{-1}(x) = x - c $. Division undoes multiplication: $latex f(x) = cx $, $latex f^{-1}(x) = \frac{x}{c} $, $latex c \neq 0 $. Example 1: Verifying Inverses Prove $latex f(x) = 2x^3 - 1 $ and $latex g(x) = \sqrt[3]{\frac{x + 1}{2}} $ are inverses: $latex f(g(x)) = x $ and $latex g(f(x)) = x $, confirming they are inverses. Theorem 1: Reflective Property If $latex f $ has the point $latex (a, b) $, then $latex f^{-1} $ contains $latex (b, a) $. Theorem 2: Existence of an Inverse A function has an inverse if and only if it is one-to-one. If $latex f $ is strictly monotonic, it is one-to-one and has an inverse. Example 2: Inverse Function Existence For $latex f(x) = x^3 + x - 1 $, since $latex f'(x) = 3x^2 + 1 > 0 $, $latex f $ is strictly monotonic and has an inverse. For $latex f(x) = x^3 - x + 1 $, it fails the Horizontal Line Test, so no inverse exists. Steps to Find an Inverse Function Check if $latex f $ is one-to-one using Theorem 2. Solve for $latex x $ in terms of $latex y $: $latex x = f^{-1}(y) $. Interchange $latex x $ and $latex y $ to find $latex y = f^{-1}(x) $. Define the domain of $latex f^{-1} $. Verify $latex f(f^{-1}(x)) = x $ and $latex f^{-1}(f(x)) = x $. Example 3: Finding Inverses For $latex f(x) = \sqrt{2x - 3} $, solve for $latex x $ in terms of $latex y $ to get $latex f^{-1}(x) = \frac{x^2 + 3}{2} $. Solution: From the graph of $latex f $, it appears that $latex f $ is increasing over its entire domain, $latex [3/2, \infty) $. So, $latex f $ is strictly monotonic, and it must have an inverse function. To find an equation for the inverse function, let $latex y = f(x) $, and solve for $latex x $ in terms of $latex y $: $latex \sqrt{2x - 3} = y $ Let y = f(x) $latex 2x - 3 = y^2 $ Square each side $latex x = \frac{y^2 + 3}{2} $ Solve for x $latex y = \frac{x^2 + 3}{2} $ Interchange x and y $latex f^{-1}(x) = \frac{x^2 + 3}{2} $ Replace y by f-1(x) The domain of $latex f^{-1} $ is the range of $latex f $, which is $latex [0, \infty) $. You can verify this result as shown below: $latex f(f^{-1}(x)) = \sqrt{2 \left( \frac{x^2 + 3}{2} \right) - 3} = \sqrt{x^2} = x, \quad x \geq 0 $ $latex f^{-1}(f(x)) = \left( \frac{\sqrt{2x - 3}}{2} \right)^2 + 3 = \frac{2x - 3 + 3}{2} = x, \quad x \geq \frac{3}{2} $ Example 4: Testing One-to-One For $latex f(x) = \sin x $, it's not one-to-one on $latex (-\infty, \infty) $, but is one-to-one on $latex [-\pi/2, \pi/2] $. Example 5 If $latex f(1) = 5 $, $latex f(3) = 7 $, and $latex f(8) = -10 $, find $latex f^{-1}(5) $, $latex f^{-1}(7) $, and $latex f^{-1}(-10) $. Solution: From the definition of $latex f^{-1} $, we have: $latex f^{-1}(5) = 1 $ because $latex f(1) = 5 $ $latex f^{-1}(7) = 3 $ because $latex f(3) = 7 $ $latex f^{-1}(-10) = 8 $ because $latex f(8) = -10 $ This shows how $latex f^{-1} $ reverses the effect of $latex f $ in this case.

Category: Math

AP Precalculus – Inverse Functions and Everything You Need to Know

Do I need AP Precalculus?