Problem 25 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ f(x)=\sqrt{|x|} ;(-1,2) $$

Step-by-Step Solution

Answer

The minimum value is 0 at \( x = 0 \). Maximum values of 1 and \( \sqrt{2} \) as \( x \to -1^+ \) and \( x \to 2^- \).

Step 1: Understand the function and interval

The function given is defined as \( f(x) = \sqrt{|x|} \). The interval provided is \((-1, 2)\), which means we need to consider values of \( x \) from just above -1 to just below 2 to find extreme values.

Step 2: Check for continuity and differentiability

The function \( f(x) = \sqrt{|x|} \) is continuous on \((-1, 2)\). However, \( f(x) \) is not differentiable at \( x = 0 \) because the absolute value inside the square root causes a kink in the graph at \( x = 0 \). This point needs special attention when checking for extreme values.

Step 3: Find critical points

To find the critical points, analyze the left and right derivative of \( f(x) \) near zero:- For \( x > 0 \), \( f(x) = \sqrt{x} \), derivative \( f'(x) = \frac{1}{2\sqrt{x}} \).- For \( x < 0 \), \( f(x) = \sqrt{-x} \), derivative \( f'(x) = -\frac{1}{2\sqrt{-x}} \).Since \( f'(0) \) does not exist, \( x = 0 \) is a critical point. Other critical points are found where \( f'(x) = 0 \), but the derivatives are non-zero for \( x eq 0 \).

Step 4: Evaluate the function at critical points and endpoints

Evaluate \( f(x) \) at critical point \( x = 0 \) and consider values approaching endpoints:- \( f(0) = \sqrt{|0|} = 0 \).- As \( x \to -1^+ \), \( f(x) \to 1 \).- As \( x \to 2^- \), \( f(x) \to \sqrt{2} \).The critical point \( x = 0 \) provides the minimum value of \( f(x) = 0 \). At \( x \to -1^+ \) and \( x \to 2^- \), the function attains maximum values of \( 1 \) and \( \sqrt{2} \) respectively.

Step 5: Conclusion for extreme values

The minimum value of the function is \( 0 \) occurring at \( x = 0 \). The function tends to its maximum values at the endpoints of the interval: \( f(x) \to 1 \) as \( x \to -1^+ \) and \( f(x) \to \sqrt{2} \) as \( x \to 2^- \). Thus, the minimum value occurs at \( x = 0 \), and the maximum values of \( 1 \) and \( \sqrt{2} \) occur at limits approaching the endpoints.

Key Concepts

Critical Points

Critical points in a function are where the derivative is either zero or does not exist. These points are important because they could be locations where a function has a local maximum or minimum. In this exercise, we looked at the function \( f(x) = \sqrt{|x|} \) in the interval \((-1, 2)\). Here, we determined a critical point at \( x = 0 \) because the derivative does not exist due to a kink in the graph caused by the absolute value function.

• For \( x > 0 \), the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) is nonzero.
• For \( x < 0 \), the derivative \( f'(x) = -\frac{1}{2\sqrt{-x}} \) is also nonzero.

Since \( f'(0) \) does not exist, \( x = 0 \) is our critical point for finding extreme values in the interval.

Continuity and Differentiability

Continuity and differentiability are foundational concepts in calculus. A function that is continuous over an interval has no gaps, jumps, or holes within the interval. Differentiability means that a function has a derivative at every point in the interval, implying a smooth graph without sharp turns.For the function \( f(x) = \sqrt{|x|} \) on the interval \((-1, 2)\):

• The function is continuous everywhere within the interval. This means it smoothly connects points without any interruptions.
• However, the function is not differentiable at \( x = 0 \) because its absolute value causes a sharp corner or kink in the graph.

When analyzing extreme values, it's crucial to examine places where the function lacks differentiability, as these could be potential sites of local maxima or minima.

Interval Analysis

Interval analysis involves evaluating a function over a specific range, in this case, \((-1, 2)\). Working within a defined interval helps in determining behavior like maximum and minimum values.With the function \( f(x) = \sqrt{|x|} \), we explored how it behaves as \( x \) approaches both endpoints of the interval:

• As \( x \to -1^+ \), the function value approaches 1.
• As \( x \to 2^- \), the function value approaches \( \sqrt{2} \).

The way a function behaves near the ends of a given interval is key in finding its extreme values. These endpoints help determine where the function reaches its maximum or minimum within the specified range.

Differentiation

Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to an independent variable. Understanding differentiation is crucial for identifying critical points, where potential extreme values can occur.For \( f(x) = \sqrt{|x|} \):

• For \( x > 0 \), it simply becomes \( f(x) = \sqrt{x} \) with a derivative of \( \frac{1}{2\sqrt{x}} \).
• For \( x < 0 \), it changes to \( f(x) = \sqrt{-x} \) with a derivative of \( -\frac{1}{2\sqrt{-x}} \).

These derivatives tell us about the slope of the function at different points in the range. A slope that doesn’t exist, like at \( x = 0 \), signals a critical point or a potential site for an extreme value, highlighting where further analysis is necessary.