Problem 5 - Brief Applied Calculus

Question

Use the definition of the derivative to show that the following functions are not differentiable at \(x=0\). \(f(x)=|2 x|\)

Step-by-Step Solution

Answer

The function is not differentiable at \( x = 0 \) because the one-sided limits of the derivative are not equal.

Step 1: Understand the function

The function given is \( f(x) = |2x| \). This function takes any input \( x \) and returns the absolute value of \( 2x \). Essentially, \( f(x) \) is equal to \( 2x \) for \( x \geq 0 \) and \( -2x \) for \( x < 0 \).

Step 2: State the definition of the derivative

The derivative of a function \( f \) at a point \( a \) is defined by \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \] if this limit exists.

Step 3: Apply the definition at \( x=0 \)

Substitute \( a = 0 \) into the derivative definition: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] Since \( f(0) = |2 \cdot 0| = 0 \), this becomes:\[ f'(0) = \lim_{h \to 0} \frac{|2h|}{h} \]

Step 4: Evaluate the limit from the right

Consider \( h > 0 \). Then \( |2h| = 2h \), and the expression becomes:\[ \lim_{h \to 0^+} \frac{2h}{h} = \lim_{h \to 0^+} 2 = 2 \]

Step 5: Evaluate the limit from the left

Consider \( h < 0 \). Then \( |2h| = -2h \), and the expression becomes:\[ \lim_{h \to 0^-} \frac{-2h}{h} = \lim_{h \to 0^-} -2 = -2 \]

Step 6: Compare the one-sided limits

The right-hand limit as \( h \to 0^+ \) is 2, and the left-hand limit as \( h \to 0^- \) is -2. Since these two one-sided limits are not equal, the two-sided limit does not exist.

Step 7: Conclusion

Since the two-sided derivative limit does not exist at \( x = 0 \), the function \( f(x) = |2x| \) is not differentiable at \( x = 0 \).

Key Concepts

Differentiability

Differentiability is a concept that helps us determine if we can find the derivative of a function at a specific point. If a function is differentiable at a point, it means we can find a specific rate of change at that position. For a function to be differentiable at a point, the derivative must exist there.

In mathematical terms, the derivative at a point \( a \) is found using the formula:

• \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)

For the function to be differentiable, the limit must exist and be the same from both directions—approaching from the left and the right.

For the function \( f(x) = |2x| \), it is not differentiable at \( x = 0 \). When we tried to find the derivative, the limit from the right at zero was 2, and from the left was -2, which shows the limit is not the same from both sides. Thus, the function fails to be differentiable at this point.

Limit

The concept of limits is central to understanding derivatives and differentiability. A limit examines what value a function approaches as the input gets closer to a particular point. When calculating derivatives, we often explore the limit of the difference quotient as \( h \) approaches zero.

In our exercise, we look at:

• Right-hand limit: \( \lim_{h \to 0^+} \)
• Left-hand limit: \( \lim_{h \to 0^-} \)

For differentiability at a point, both these limits must not only exist but also be equal. If they differ, the limit does not exist in a way that allows for differentiability. As seen with \( f(x) = |2x| \) at \( x = 0 \), the limits aren't equal: 2 from the right, and -2 from the left. Hence, it's not differentiable.

Absolute Value

The absolute value function is fundamental in this problem, defined as the non-negative value of a number without regard to its sign. For \( f(x) = |2x| \), the absolute value determines how the function behaves:

• For \( x \geq 0 \), \( f(x) = 2x \)
• For \( x < 0 \), \( f(x) = -2x \)

Because absolute value influences the way a function is structured—splitting its definition based on the sign of \( x \)—it can lead to situations where the function has different rates of change on either side of a point. This was the case at \( x = 0 \); the behavior of \( |2x| \) led to differing one-sided limits. This makes the function non-smooth at the point and hence non-differentiable there.