Question

If \(f(x)=\left\\{\begin{array}{cl}x & , 0 \leq x \leq 1 \\ 2-e^{x-1}, & 1

Short answer

Option (3) is correct; \g(x)\ has local maxima at \x = 1 + \text{ln}(2)\ and local minima at \x = e\.

Step-by-step solution

Define the Problem

Given is a piecewise function for \(f(x)\) and relationship \(g^{'}(x) = f(x)\). You need to identify whether \(g(x)\) has local maxima and/or minima.

Understand the Relationship Between g(x) and f(x)

Since \(g^{'}(x) = f(x)\), finding the critical points of \(g(x)\) requires setting \(f(x) = 0\) and solving for \(x\).

Solve for Critical Points

Examine each piece of \(f(x)\) from the given intervals to determine where \(f(x) = 0\). Specifically look at \[ f(x) = 2 - e^{x-1} \] and \[ f(x) = x - e \] since these sections may contain solutions in the given range.

Finding Zeroes for Each Piece

For the interval \(1 < x \leq 2\), solve \(2 - e^{x-1} = 0\).This gives \(e^{x-1} = 2\), and solving it gives \(x = 1 + \text{ln}(2)\).For the interval \(2 < x \leq 3\), solve \(x - e = 0\).This gives \(x = e\).

Analyze the Nature of Critical Points

To determine whether each critical point is a local minimum or maximum, analyze the behavior of \(f(x)\) around the critical points.At \(x = 1 + \text{ln}(2)\), check the sign change of \(f(x)\) around this point.Similarly at \(x = e\), check the sign changes.

Determine the Nature of Critical Points

- For \(x = 1 + \text{ln}(2)\):\(f(x) = 2 - e^{x-1}\).If \(x \lt 1 + \text{ln}(2)\), \(f(x) \gt 0\);if \(x \gt 1 + \text{ln}(2)\), \(f(x) \lt 0\).Hence, \x = 1 + \text{ln}(2)\ is a local maximum.- For \(x = e\):\(f(x) = x - e\).If \(x \lt e\), \(f(x) \lt 0\);if \(x \gt e\), \(f(x) \gt 0\).Hence, \x = e\ is a local minimum.

Key concepts

critical points

To understand the behavior of functions like \(g(x)\), we must find its critical points. Critical points are values of \(x\) where the derivative of the function, \(g'(x)\), is zero or undefined. These points often indicate where the function has a local maximum, local minimum, or a horizontal tangent.
For the given problem, \(g'(x) = f(x)\). Therefore, we need to find where \(f(x) = 0\).
From the piecewise function definition:

• \f(x) = 2 - e^{x-1}\ for \(1 < x \leq 2\)
• \f(x) = x - e\ for \(2 < x \leq 3\)

We need to solve these parts individually for \(f(x) = 0\).
For \(2 - e^{x-1} = 0\), solving gives \(x = 1 + \text{ln}(2)\).
For \(x - e = 0\), solving gives \(x = e\). Thus, the critical points are \(x = 1 + \text{ln}(2)\) and \(x = e\).

piecewise function analysis

Piecewise functions have separate definitions over different intervals. This means we need to treat each piece of the function individually instead of as a whole. For our given problem, the piecewise function \(f(x)\) has different equations over different intervals:
1. \(f(x) = x\) for \(0 \leq x \leq 1\)
2. \(f(x) = 2 - e^{x-1}\) for \(1 < x \leq 2\)
3. \(f(x) = x - e\) for \(2 < x \leq 3\)
When solving for critical points, or examining behavior, you must focus on one piece at a time.
For example, to check where \(f(x) = 0\) in the interval \(1 < x \leq 2\), solve \(2 - e^{x-1} = 0\), which simplifies to \(e^{x-1} = 2\) giving \(x = 1 + \text{ln}(2)\). Similarly, solve for the piece over \(2 < x \leq 3\), which provides \(x = e\). This method ensures you accurately understand the function's overall behavior.

sign change in derivatives

Determining whether critical points are maxima or minima involves checking the sign change of the derivative around these points. For our given function, we check the sign change of \(f(x)\), the same as \(g'(x)\), at the critical points.
Consider \(x = 1 + \text{ln}(2)\) in the interval \(1 < x \leq 2\). The function is \(f(x) = 2 - e^{x-1}\).

• If \(x < 1 + \text{ln}(2)\), \(f(x) > 0\)
• If \(x > 1 + \text{ln}(2)\), \(f(x) < 0\)

This indicates \(x = 1 + \text{ln}(2)\) is a local maximum.
Next, consider \(x = e\) in the interval \(2 < x \leq 3\). The function is \(f(x) = x - e\).

• If \(x < e\), \(f(x) < 0\)
• If \(x > e\), \(f(x) > 0\)

This indicates \(x = e\) is a local minimum.
Checking sign changes in derivatives helps determine the nature of the critical points accurately.