Math Problem Statement

Show that f(x) = |x| + |x - 1| is not differentiable at x = 1.

Solution

The image shows a question related to differentiability. Let me summarize and analyze it:

The question seems to be asking to show that the function: f(x)=x+x1f(x) = |x| + |x - 1| is not differentiable at x=1x = 1.

To determine if f(x)f(x) is differentiable at x=1x = 1, we can check the behavior of the derivative from the left and right of this point. The function f(x)=x+x1f(x) = |x| + |x - 1| involves absolute values, which typically leads to points where differentiability may fail, especially at points where the argument of the absolute value changes sign (here, x=0x = 0 and x=1x = 1).

Steps to Determine Differentiability at x=1x = 1

  1. Consider the behavior for x<1x < 1:

    • For x<1x < 1, x=x|x| = x if x0x \ge 0 and x=x|x| = -x if x<0x < 0.
    • x1=1x|x - 1| = 1 - x.

    This gives two cases to consider within x<1x < 1:

    • If 0x<10 \le x < 1: f(x)=x+(1x)=1f(x) = x + (1 - x) = 1.
    • If x<0x < 0: f(x)=x+(1x)=12xf(x) = -x + (1 - x) = 1 - 2x.
  2. For x=1x = 1:

    • We can calculate the left-hand and right-hand limits of the derivative at x=1x = 1.
  3. Calculate the derivatives on either side of x=1x = 1:

    • For x>1x > 1, f(x)=x+(x1)=2x1f(x) = x + (x - 1) = 2x - 1, which has a derivative of 22.
    • For x<1x < 1 but x0x \ge 0, f(x)f(x) is constant at f(x)=1f(x) = 1, so the derivative is 00.
  4. Conclusion:

    • The left-hand derivative at x=1x = 1 is 00, and the right-hand derivative is 22.
    • Since the left-hand and right-hand derivatives do not match, f(x)f(x) is not differentiable at x=1x = 1.

Would you like a detailed step-by-step breakdown, or do you have further questions?

Related Questions:

  1. How do we evaluate the differentiability of a function with absolute values?
  2. What are other methods to check differentiability at a given point?
  3. Why does the presence of an absolute value function often lead to non-differentiability?
  4. Can you find points of non-differentiability for f(x)=x+x1f(x) = |x| + |x - 1| other than x=1x = 1?
  5. How does the behavior of the derivative near x=1x = 1 help in determining differentiability?

Tip:

For functions involving absolute values, consider breaking down the function into cases based on the intervals where the expressions inside the absolute values are positive or negative.

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Math Problem Analysis

Mathematical Concepts

Differentiability
Absolute Value Function
Piecewise Functions

Formulas

f(x) = |x| + |x - 1|

Theorems

Definition of Differentiability
Left-hand and Right-hand Derivatives

Suitable Grade Level

Grades 11-12