Given \(y = f(x) = x^3\), find the value of input when output is given:
Solution:
Inverse Function: If \(g(x) = \sqrt[3]{x}\) is the inverse function for \(f(x) = x^3\), then:
Notation: \(g(x) = f^{-1}(x)\)
For \(f(x) = x^2\), find input when output = 4:
Issue: Can be 2 or -2
Let output = \(y\), so input = \(\sqrt{y}\) or \(-\sqrt{y}\)
Conclusion: Inverse does not exist for \(f(x) = x^2\) on all real numbers.
Condition for Inverse to Exist: When the function is one-to-one (1-1) in its domain.
One-to-One Definition: No two inputs have the same output.
Graph Analysis: The handwritten notes show two graphs side by side. The left graph shows y = x² (a parabola) with horizontal lines intersecting it at multiple points, demonstrating it fails the horizontal line test. The right graph shows y = x³ with horizontal lines each intersecting at only one point, showing it passes the test.
Horizontal Line Test: Used to determine whether a given graph is 1-1 or not.
If every horizontal line intersects the graph at most once, the function is 1-1 and has an inverse.
Results from the Test:
Graph Illustration: The notes show the parabola y = x² with the right half highlighted, restricting the domain to [0,∞) to make it one-to-one.
Original Domain Restriction: \(D: [0, \infty)\), \(R: [0, \infty)\)
When \(y\) is output, input = \(\sqrt{y}\)
\(f^{-1}(y) = \sqrt{y}\) or \(f^{-1}(x) = \sqrt{x}\)
Inverse Function: \(D: [0, \infty)\), \(R: [0, \infty)\)
Alternative Restriction: \(D: (-\infty, 0]\), \(R: [0, \infty)\)
When \(y\) is output, input = \(-\sqrt{y}\)
\(f^{-1}(y) = -\sqrt{y}\) or \(f^{-1}(x) = -\sqrt{x}\)
Inverse Function: \(D: [0, \infty)\), \(R: (-\infty, 0]\)
For \(f(x) = |x-5|\):
Graph Features: The notes show a V-shaped graph representing the absolute value function, with the vertex at (5,0) and both sides of the V highlighted in different colors to show possible domain restrictions.
Function is not invertible on \((-\infty, \infty)\)
Option 1 - Restrict to right side: \(D: [5, \infty)\), \(R: [0, \infty)\)
For \(x \geq 5\): \(y = x - 5\), so \(x = y + 5\)
\(f^{-1}(y) = y + 5\) or \(f^{-1}(x) = x + 5\)
Option 2 - Restrict to left side: \(D: (-\infty, 5]\), \(R: [0, \infty)\)
For \(x \leq 5\): \(y = 5 - x\), so \(x = 5 - y\)
\(f^{-1}(y) = 5 - y\) or \(f^{-1}(x) = 5 - x\)
For \(f(x) = \frac{1}{x-3}\), find \(f^{-1}(x)\):
Graph Characteristics: The notes show a hyperbola with vertical asymptote at x = 3 and horizontal asymptote at y = 0, passing the horizontal line test and therefore having an inverse.
Solution:
The function is 1-1 in the domain \(D: (-\infty, 3) \cup (3, \infty)\)
Finding inverse: \(y = \frac{1}{x-3}\)
Solve for \(x\): \(x - 3 = \frac{1}{y}\), so \(x = \frac{1}{y} + 3\)
\(f^{-1}(y) = \frac{1}{y} + 3\) or \(f^{-1}(x) = \frac{1}{x} + 3\)
Domain of \(f^{-1}\): \((-\infty, 0) \cup (0, \infty)\)
Range of \(f^{-1}\): \((-\infty, 3) \cup (3, \infty)\)
Graph Relationship: The handwritten notes show two graphs side by side - an exponential function y = b^x and its inverse the logarithmic function, illustrating how they are reflections of each other across the line y = x.
Exponential Function: \(y = b^x\) where \(b > 0\) and \(b \neq 1\)
Domain: \((-\infty, \infty)\), Range: \((0, \infty)\)
Logarithmic Function: The inverse of \(y = b^x\) is the logarithmic function
\(f^{-1}(x) = \log_b x\)
Domain: \((0, \infty)\), Range: \((-\infty, \infty)\)
Fundamental Properties:
Logarithm Rules:
Find \(x\) such that \(3^{x-2} = 18\):
Solution:
Take \(\log_3\) of both sides:
\[\log_3(3^{x-2}) = \log_3(18)\] \[x-2 = \log_3(18)\]Using logarithm properties:
\[\log_3(18) = \log_3(3^2 \cdot 2) = \log_3(3^2) + \log_3(2) = 2 + \log_3(2)\]Therefore:
\[x - 2 = 2 + \log_3(2)\] \[x = \frac{1}{2}(4 + \log_3(2))\]Write \(6^x\) as exponential with base \(e\):
Solution:
For any \(y\): \(e^{\ln y} = y\)
Let \(y = 6^x\):
\[e^{\ln(6^x)} = 6^x\] \[e^{x \ln 6} = 6^x\]Key Relationship: The exponential and logarithmic functions are inverses of each other. The domain of one becomes the range of the other, and their graphs are reflections across the line \(y = x\).