Question

Let $f(x)=10-|x-5|,\; x\in\mathbb{R}$. Then $f(x)$ is not differentiable at:

• A. $x=10$
• B. $x=15$
• C. $x=-5$
• D. $x=5$
• E. $x=-15$

Hint:

To find non-differentiable points for absolute value functions, set the expression inside the absolute value to zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Differentiability of a function at a point requires the function to be continuous and have a unique, non-vertical tangent at that point. Functions involving absolute values often have "sharp corners" or "cusps" where the derivative is undefined.
Step 2: Key Formula or Approach:
The absolute value function \( g(u) = |u| \) is not differentiable at the point where its argument is zero, i.e., at \( u=0 \).
For our function \( f(x) = 10 - |x-5| \), the non-differentiable point will occur where the argument of the absolute value, \( x-5 \), is equal to zero.
Step 3: Detailed Explanation:
The function is \( f(x) = 10 - |x-5| \).
The part of the function that causes a potential issue with differentiability is \( |x-5| \).
The function \( |x-5| \) can be defined piecewise:
\[ |x-5| = \begin{cases} x-5 & \text{if } x-5 \geq 0 \implies x \geq 5
-(x-5) & \text{if } x-5<0 \implies x<5 \end{cases} \] This means our function \( f(x) \) is also piecewise:
\[ f(x) = \begin{cases} 10 - (x-5) = 15 - x & \text{if } x \geq 5
10 - (-(x-5)) = 10 + x - 5 = 5 + x & \text{if } x<5 \end{cases} \] To check for differentiability at x=5, we need to compare the left-hand derivative and the right-hand derivative at that point.
Right-hand derivative (for \( x \geq 5 \)):
\[ f'(x) = \frac{d}{dx}(15-x) = -1 \] So, \( f'_+(5) = -1 \).
Left-hand derivative (for \( x<5 \)):
\[ f'(x) = \frac{d}{dx}(5+x) = 1 \] So, \( f'_-(5) = 1 \).
Since the left-hand derivative (1) is not equal to the right-hand derivative (-1) at x=5, the function is not differentiable at x=5. This point corresponds to a sharp corner on the graph of the function.
Step 4: Final Answer:
The function \( f(x) \) is not differentiable at x=5.