Chapter 3: Problem 33
Exer. \(33-36:\) Use right-hand and left-hand derivatives to prove that \(f\) is not differentiable at \(a\). $$ f(x)=|x-5| ; \quad a=5 $$
Short Answer
Expert verified
The function is not differentiable at \(x = 5\) because the left-hand and right-hand derivatives are not equal.
Step by step solution
01
Understanding the Function
The function given is \( f(x) = |x - 5| \). The absolute value function creates a V-shape graph with a vertex at \(x = 5\). We'll examine if the function is differentiable at \(a = 5\).
02
Definition of Differentiability
A function is differentiable at a point if the left-hand derivative and right-hand derivative at that point exist and are equal. We need to compute these derivatives for \(x = 5\).
03
Calculating the Left-Hand Derivative
For \(x < 5\), \(f(x) = 5 - x\). The left-hand derivative at \(x = 5\) is computed as: \[ \lim_{h \to 0^-} \frac{f(5 + h) - f(5)}{h} = \lim_{h \to 0^-} \frac{5 - (5 + h)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1. \] This gives us the left-hand derivative as \(-1\).
04
Calculating the Right-Hand Derivative
For \(x > 5\), \(f(x) = x - 5\). The right-hand derivative at \(x = 5\) is computed as: \[ \lim_{h \to 0^+} \frac{f(5 + h) - f(5)}{h} = \lim_{h \to 0^+} \frac{(5 + h) - 5}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1. \] This gives us the right-hand derivative as \(1\).
05
Conclusion on Differentiability
Since the left-hand derivative \(-1\) is not equal to the right-hand derivative \(1\), \(f(x)\) is not differentiable at \(x = 5\). At \(x = 5\), the function has a sharp corner.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Function
The absolute value function is a crucial mathematical tool that measures the "magnitude" of a number without considering its sign. In simpler terms, it makes any negative quantity positive. In our exercise, the function in question is defined as \( f(x) = |x - 5| \). This particular format shows how the function transforms depending on the value of \( x \) in relation to 5.
- If \( x < 5 \), the expression inside the absolute value, \( x - 5 \), becomes negative, and as a result, \( f(x) = 5 - x \).
- If \( x > 5 \), the expression inside is positive, so \( f(x) = x - 5 \).
Left-Hand Derivative
The left-hand derivative is a mathematical concept that describes the rate of change of a function as you approach a point from the left. Essentially, it's a way of capturing how the function behaves just before reaching the specific point.
To calculate the left-hand derivative of \( f(x) = |x - 5| \) at \( x = 5 \), we use the fact that when \( x < 5 \), \( f(x) = 5 - x \). We compute:\[\lim_{h \to 0^-} \frac{f(5 + h) - f(5)}{h}\]Substituting, we get the expression:\[\lim_{h \to 0^-} \frac{5 - (5 + h)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\]This step confirms that the left-hand derivative of the function at \( x = 5 \) is \(-1\). This result is critical because differentiability requires both left and right derivatives to exist and to be equal.
To calculate the left-hand derivative of \( f(x) = |x - 5| \) at \( x = 5 \), we use the fact that when \( x < 5 \), \( f(x) = 5 - x \). We compute:\[\lim_{h \to 0^-} \frac{f(5 + h) - f(5)}{h}\]Substituting, we get the expression:\[\lim_{h \to 0^-} \frac{5 - (5 + h)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\]This step confirms that the left-hand derivative of the function at \( x = 5 \) is \(-1\). This result is critical because differentiability requires both left and right derivatives to exist and to be equal.
Right-Hand Derivative
The right-hand derivative operates similarly to the left-hand derivative but assesses the rate of change as you approach a point from the right. It tends to emphasize how the function is behaving immediately after the specific point under consideration.
For the function \( f(x) = |x - 5| \), we need to compute this derivative at \( x = 5 \) by considering \( f(x) = x - 5 \) for \( x > 5 \). Our calculation is:\[\lim_{h \to 0^+} \frac{f(5 + h) - f(5)}{h}\]Plugging in the values gives:\[\lim_{h \to 0^+} \frac{(5 + h) - 5}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\]This result shows that the right-hand derivative of the function at \( x = 5 \) is \(1\). As with the left-hand side, identifying this derivative is necessary to determine the differentiability of the function.
For the function \( f(x) = |x - 5| \), we need to compute this derivative at \( x = 5 \) by considering \( f(x) = x - 5 \) for \( x > 5 \). Our calculation is:\[\lim_{h \to 0^+} \frac{f(5 + h) - f(5)}{h}\]Plugging in the values gives:\[\lim_{h \to 0^+} \frac{(5 + h) - 5}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\]This result shows that the right-hand derivative of the function at \( x = 5 \) is \(1\). As with the left-hand side, identifying this derivative is necessary to determine the differentiability of the function.
Sharp Corner
A sharp corner is a point on a graph where there is a distinct change in the direction of the curve, often resulting in an instantaneous and dramatic shift. In the context of differentiability, sharp corners present a unique challenge.
At \( x = 5 \) in the function \( f(x) = |x - 5| \), we observe such a corner. From our previous findings:
At \( x = 5 \) in the function \( f(x) = |x - 5| \), we observe such a corner. From our previous findings:
- The left-hand derivative is \(-1\).
- The right-hand derivative is \(1\).