AP Calculus BC 3.3 Differentiating Inverse Functions Study Notes - New Syllabus
AP Calculus BC 3.3 Differentiating Inverse Functions Study Notes- New syllabus
AP Calculus BC 3.3 Differentiating Inverse Functions Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Calculate derivatives of Inverse functions.
Key Concepts:
- Differentiating Inverse Functions
Derivatives of Inverse Functions
Inverse Function
The inverse of a function reverses the effect of the original function. If \( f(x) \) is a one-to-one function, its inverse is denoted as \( f^{-1}(x) \) and satisfies:
\( f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \).
Condition for Inverse:
- The function \( f \) must be one-to-one (injective), meaning no two different inputs produce the same output.
- This is equivalent to passing the Horizontal Line Test on its graph.
Graphical Interpretation:
The graph of \( f^{-1}(x) \) is the reflection of the graph of \( f(x) \) across the line \( y = x \).
Derivatives of Inverse Functions
If \( y = f(x) \) is differentiable and one-to-one, then its inverse \( x = f^{-1}(y) \) satisfies:
\( \dfrac{d}{dx}[f^{-1}(x)] = \dfrac{1}{f'(f^{-1}(x))} \).
Example:
Find \( (f^{-1})'(3) \) if \( f(x) = x^3 + x \).
▶️ Answer/Explanation
Step 1: Formula: \( (f^{-1})'(a) = \dfrac{1}{f'(b)} \), where \( f(b) = a \).
Step 2: Find \( b \) such that \( f(b) = 3 \):
\( x^3 + x = 3 \).
Try \( x = 1 \): \( 1^3 + 1 = 2 \) (too small).
Try \( x = 1.2 \): \( (1.2)^3 + 1.2 = 1.728 + 1.2 = 2.928 \approx 3 \).
So \( b \approx 1.2 \).
Step 3: Compute derivative: \( f'(x) = 3x^2 + 1 \).
\( f'(1.2) = 3(1.44) + 1 = 4.32 + 1 = 5.32 \).
Step 4: Apply formula:
\( (f^{-1})'(3) \approx \dfrac{1}{5.32} \approx 0.188 \).
Answer: Approximately \( 0.188 \).
Example:
If \( f(x) = \ln(x) \), find \( (f^{-1})'(5) \).
▶️ Answer/Explanation
Step 1: The inverse of \( \ln(x) \) is \( e^x \).
Step 2: Formula: \( (f^{-1})'(a) = \dfrac{1}{f'(b)} \), where \( f(b) = a \).
\( f(b) = a \Rightarrow \ln(b) = 5 \Rightarrow b = e^5 \).
Step 3: Compute derivative: \( f'(x) = \dfrac{1}{x} \).
\( f'(e^5) = \dfrac{1}{e^5} \).
Step 4: Apply formula:
\( (f^{-1})'(5) = \dfrac{1}{1/e^5} = e^5 \).
Answer: \( e^5 \).
Example:
Find \( (f^{-1})'(a) \) given \( f(x) = x^5 – x^3 + 2x \) and \( a = 2 \).
▶️ Answer/Explanation
Use the inverse derivative formula:
\( (f^{-1})'(a) = \dfrac{1}{f'(b)} \), where \( b = f^{-1}(a) \) (so \( f(b) = a \)).
Find \( b \) such that \( f(b) = 2 \):
\( f(x) = x^5 – x^3 + 2x \)
\( 2 = x^5 – x^3 + 2x \)
Trial gives \( x = 1 \). So \( b = 1 \).
\( f'(x) = 5x^4 – 3x^2 + 2 \).
Plug in \( x = 1 \):
\( f'(1) = 5(1)^4 – 3(1)^2 + 2 = 5 – 3 + 2 = 4 \).
Compute \( (f^{-1})'(2) \):
\( (f^{-1})'(2) = \dfrac{1}{4} \).