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Show that \(f(x)=|x-1|\) has a local minimum at \(x=1\) but \(f(x)\) is not differentiable at \(x=1\).

Short Answer

Expert verified
\( f(x) \) has a local minimum at \( x=1 \) but is not differentiable there.

Step by step solution

01

Understand the Function

The function given is \( f(x) = |x-1| \). This is an absolute value function, which means it creates a V-shape graph where the vertex is located at the point where the inside expression is zero.
02

Determine the Vertex

Set \( x-1 = 0 \) to find the x-coordinate of the vertex. Solving gives \( x = 1 \). At \( x = 1 \), \( f(x) = |1-1| = 0 \). Thus, the vertex, or the point of interest, is at (1, 0).
03

Analyze the Graph Shape

Since an absolute value function \( |x-c| \) forms a V-shape, the vertex at \( x = 1 \) is the lowest point. As such, \( f(x) \) has a local minimum at \( x = 1 \).
04

Check Differentiability

For differentiability at \( x = 1 \), check the left and right-hand derivatives. For \( x < 1 \), \( f(x) = -(x-1) = 1-x \), and its derivative is \( f'(x) = -1 \). For \( x > 1 \), \( f(x) = x-1 \), and its derivative is \( f'(x) = 1 \). These one-sided derivatives (-1 and 1) are not equal at \( x = 1 \), thus \( f(x) \) is not differentiable there.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Local Minimum
An absolute value function like \( f(x) = |x - 1| \) often features a local minimum due to its V-shaped graph. In mathematics, a local minimum is a point where the function value is lower than at any other nearby points. For the function \( f(x) = |x - 1| \), the local minimum occurs at the vertex of the V-shape. This is where the expression inside the absolute value equals zero, which we find by setting \( x - 1 = 0 \), resulting in \( x = 1 \).

At \( x = 1 \), the value of the function is minimized to \( f(1) = |1-1| = 0 \). This tells us that the point \( (1, 0) \) is the lowest point on the graph, indicating a local minimum. This attribute of absolute value functions ensures that their vertices often serve as local minima, making them straightforward to identify.

In practical terms, identifying a local minimum is essential when analyzing trends or optimizing functions in various applications like physics or economics. It represents a stable point where small changes to input \( x \) result in no decrease in the value of \( f(x) \).
Differentiability
Differentiability refers to the ability of a function to have a derivative at a particular point. For the absolute value function \( f(x) = |x - 1| \), checking differentiability at \( x = 1 \) involves looking at the behavior of the slope on either side of this point.

On each side of \( x = 1 \), the function behaves differently due to the nature of absolute values:
  • For \( x < 1 \), the function is \( f(x) = -(x-1) = 1 - x \), with a derivative \( f'(x) = -1 \).
  • For \( x > 1 \), the function is \( f(x) = x-1 \), with a derivative \( f'(x) = 1 \).
The derivatives \( -1 \) and \( 1 \) are not equal, illustrating that there is a jump, or cusp, at \( x = 1 \). This difference means the function does not have a well-defined tangent line at that point, thus \( f(x) \) is not differentiable at \( x = 1 \).

Understanding differentiability helps in determining where functions are smooth and continuous vs. where they have sharp turns or corners. Such insights are valuable for simulations and modeling scenarios where smoothness impacts outcomes.
Vertex
In the context of the absolute value function \( f(x) = |x-1| \), the vertex is the critical point where the function's direction changes, presenting the V-shape. The vertex for this function is found by determining where the inside expression \( x - 1 \) becomes zero. By solving \( x - 1 = 0 \), the x-coordinate of the vertex is discovered to be \( x = 1 \).

At \( x = 1 \), the function value is minimized, resulting in the vertex point \( (1, 0) \). The vertex effectively indicates the transition from a decreasing to an increasing function as we move from left to right on the graph. This is why the vertex often corresponds to the local minimum for absolute value functions. It's the "turning point" of the graph.

The concept of a vertex extends beyond just visuals; it is crucial in solving real-world problems involving optimization. By identifying the vertex, one can quickly determine points of lowest or highest function values, useful for evaluating minimum costs or maximum efficiency in engineering or economics.

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Most popular questions from this chapter

Elimination of ethanol from the blood is known to have zeroth order kinetics. Provided no more ethanol enters the blood, the concentration of ethanol in a person's blood will therefore obey the following differential equation: $$ \frac{d M}{d t}=-k_{0} $$ where for a typical adult \(k_{0}=0.186 \mathrm{~g} /\) liter \(/ \mathrm{hr}\) (al-Lanqawi et al. 1992). (a) Explain why \(M(t)\) can only obey the above differential equation if \(M>0\) (once \(M\) drops to 0 , it is usual to assume that \(\left.\frac{d M}{d t}=0\right)\) (b) If a person's blood alcohol concentration is \(1.6 \mathrm{~g}\) /liter at midnight, what will their blood alcohol concentration be at \(2 \mathrm{am}\) ? You may assume that she drinks no more alcohol after midnight. (c) At what time will their blood alcohol concentration drop to \(0 \mathrm{~g} /\) liter?

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Find the equilibrium point of $$x_{t+1}=x_{t}+\frac{1}{2} e^{-x_{t}}-\frac{1}{2}, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether it is stable or unstable.

Find the general solution of the differential equation. $$ \frac{d y}{d x}=\frac{2}{x}-x, x>0 $$

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