AP Calculus BC 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist Study Notes - New Syllabus
AP Calculus BC 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist Study Notes- New syllabus
AP Calculus BC 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Explain the relationship between differentiability and continuity.
Key Concepts:
- Connecting Differentiability and Continuity
Connecting Differentiability and Continuity
Connecting Differentiability and Continuity
A function must be continuous at a point in order to be differentiable at that point. However, the reverse is not always true: a function can be continuous but not differentiable.
- If a function \( f(x) \) is differentiable at \( x = a \), then it is also continuous at \( x = a \).
- If a function \( f(x) \) is not continuous at \( x = a \), then it is not differentiable at \( x = a \).
When Derivatives Do Not Exist:
A derivative does not exist at a point where any of the following occurs:
- Discontinuity: The function is not continuous at that point.
- Corner or Cusp: The left-hand and right-hand slopes are not equal.
- Vertical Tangent: The tangent line becomes vertical, so the slope is undefined.
Graphical Signs of Non-Differentiability:
- A sharp turn or point (like \( f(x) = |x| \) at \( x = 0 \)).
- A jump or hole in the graph (discontinuity).
- A vertical tangent (like \( f(x) = \sqrt[4]{x} \)).
Example:
Consider the function:
\( f(x) = |x| \)
Sketch the graph of \( f(x) \) and determine whether it is differentiable at \( x = 0 \).
▶️Answer/Explanation
- The graph of \( f(x) = |x| \) has a sharp corner at \( x = 0 \).
- The left-hand derivative at \( x = 0 \) is \( -1 \), and the right-hand derivative is \( +1 \).
- Since these one-sided derivatives are not equal, the derivative does not exist at \( x = 0 \).
- However, the function is continuous at \( x = 0 \) because \( \lim_{x \to 0} f(x) = f(0) = 0 \).
Conclusion: \( f(x) = |x| \) is continuous but not differentiable at \( x = 0 \).
Example :
Let \( f(x) = \sqrt[3]{x} \). Is the function differentiable at \( x = 0 \)?
▶️Answer/Explanation
- The function \( f(x) = \sqrt[3]{x} \) is continuous everywhere including \( x = 0 \).
- Its derivative is \( f'(x) = \dfrac{1}{3x^{2/3}} \).
- At \( x = 0 \), the derivative is undefined because of division by zero.
- This means the graph has a vertical tangent at \( x = 0 \).
Conclusion: \( f(x) = \sqrt[3]{x} \) is continuous at \( x = 0 \) but not differentiable there due to a vertical tangent line.
Example:
Let \( f(x) = \begin{cases} 1, & x < 0 \\ 2, & x \geq 0 \end{cases} \). Is the function differentiable at \( x = 0 \)?
▶️Answer/Explanation
- This is a piecewise-defined function with a jump from 1 to 2 at \( x = 0 \).
- \( \lim_{x \to 0^-} f(x) = 1 \), but \( \lim_{x \to 0^+} f(x) = 2 \)
- Since the left and right limits are not equal, \( \lim_{x \to 0} f(x) \) does not exist.
- Therefore, the function is not continuous at \( x = 0 \), and hence not differentiable there.
Conclusion: Because of the jump discontinuity at \( x = 0 \), \( f(x) \) is neither continuous nor differentiable at that point.
Example:
Let \( f(x) = \begin{cases} x^2, & x \leq 2 \\ 4x – 4, & x > 2 \end{cases} \). Is \( f(x) \) differentiable at \( x = 2 \)?
▶️Answer/Explanation
Check continuity first:
- \( f(2) = 2^2 = 4 \)
- \( \lim_{x \to 2^-} f(x) = 4 \), \( \lim_{x \to 2^+} f(x) = 4(2) – 4 = 4 \)
- So, the function is continuous at \( x = 2 \).
Now check differentiability:
- Left derivative: \( \frac{d}{dx}(x^2) = 2x \Rightarrow 2(2) = 4 \)
- Right derivative: \( \frac{d}{dx}(4x – 4) = 4 \)
- Since both left and right derivatives are equal, the function is differentiable at \( x = 2 \).
Conclusion: \( f(x) \) is both continuous and differentiable at \( x = 2 \).