Differentiability of Piecewise Functions - Part(4)
These calculus questions focus on the differentiability of functions.
Each problem is solved step by step using fundamental theorems of continuity and derivatives.
Theorems
Theorem 1.
If a function \( f \) is differentiable at \( x = a \), then \( f \) is continuous at \( x = a \).
Contrapositive.
If \( f \) is not continuous at \( x = a \), then \( f \) is not differentiable at \( x = a \).
Theorem 2.
If \( f \) is continuous at \( x = a \) and
\[
\lim_{x \to a^+} f'(x) = \lim_{x \to a^-} f'(x),
\]
then \( f \) is differentiable at \( x = a \) and
\[
f'(a) = \lim_{x \to a^+} f'(x) = \lim_{x \to a^-} f'(x).
\]
Question 1
Determine whether the function
\[
f(x) =
\begin{cases}
2x^2, & x \le 1 \\
2\sqrt{x}, & x > 1
\end{cases}
\]
is differentiable at \( x = 1 \).
Solution
-
Evaluate continuity at \( x = 1 \):
\[
f(1) = 2(1)^2 = 2
\]
\[
\lim_{x \to 1^-} f(x) = 2, \quad
\lim_{x \to 1^+} f(x) = 2
\]
-
Since the limits equal \( f(1) \), the function is continuous at \( x = 1 \).
-
Compute derivatives:
\[
f'(x) = 4x \quad (x < 1), \qquad
f'(x) = \frac{1}{\sqrt{x}} \quad (x > 1)
\]
-
Evaluate derivative limits:
\[
\lim_{x \to 1^-} f'(x) = 4, \quad
\lim_{x \to 1^+} f'(x) = 1
\]
-
Since the limits are not equal, \( f'(1) \) does not exist.
The function is not differentiable at \( x = 1 \).
Question 2
Let
\[
f(x) =
\begin{cases}
x^3, & x \le 0 \\
x^3 + 1, & x > 0
\end{cases}
\]
Show that although
\[
\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^-} f'(x),
\]
the derivative \( f'(0) \) does not exist.
Solution
-
\[
\lim_{x \to 0^-} f(x) = 0, \quad
\lim_{x \to 0^+} f(x) = 1
\]
-
The function is not continuous at \( x = 0 \), hence not differentiable there.
-
For all \( x \neq 0 \):
\[
f'(x) = 3x^2
\]
-
\[
\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^+} f'(x) = 0
\]
-
Even though the derivative limits agree, \( f'(0) \) does not exist due to discontinuity.
Question 3
Find constants \( A \) and \( B \) such that
\[
f(x) =
\begin{cases}
2x^2, & x \le 2 \\
Ax + B, & x > 2
\end{cases}
\]
is differentiable at \( x = 2 \).
Solution
-
Continuity at \( x = 2 \):
\[
\lim_{x \to 2^-} f(x) = 8, \quad
\lim_{x \to 2^+} f(x) = 2A + B
\]
-
\[
2A + B = 8
\]
-
Derivatives:
\[
f'(x) = 4x \ (x<2), \quad f'(x) = A \ (x>2)
\]
-
\[
\lim_{x \to 2^-} f'(x) = 8, \quad
\lim_{x \to 2^+} f'(x) = A
\]
-
Thus \( A = 8 \) and \( B = -8 \).
Question 4
Find all values of \( x \) for which
\[
f(x) = \sqrt{x^2 - 2x + 1}
\]
is not differentiable.
Solution
-
\[
f(x) = \sqrt{(x-1)^2} = |x-1|
\]
-
\[
f(x) =
\begin{cases}
x - 1, & x > 1 \\
-(x - 1), & x < 1
\end{cases}
\]
-
The function is continuous at \( x = 1 \).
-
Derivatives:
\[
f'(x) = 1 \ (x<1), \quad f'(x) = -1 \ (x>1)
\]
-
The derivative limits differ, so \( f \) is not differentiable at \( x = 1 \).
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