Problem 32 - Thomas Calculus

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$ g(x)=4 \sqrt{x}-x^{2}+3 $$

Step-by-Step Solution

Answer

The function is increasing on \((0, 1)\) and decreasing on \((1, \infty)\). It has a local and absolute maximum of 6 at \(x=1\).

Step 1: Find the derivative

To find the intervals where the function is increasing or decreasing, we need the derivative of the function. Given the function \( g(x) = 4\sqrt{x} - x^2 + 3 \), we differentiate it with respect to \( x \). The derivative is: \[ g'(x) = \frac{4}{2\sqrt{x}} - 2x = \frac{2}{\sqrt{x}} - 2x \]

Step 2: Solve for critical points

Critical points occur where the derivative \( g'(x) \) is either zero or undefined. The derivative is undefined at \( x=0 \). To find where it is zero, solve: \[ \frac{2}{\sqrt{x}} = 2x \]Multiply both sides by \( \sqrt{x} \) to clear the fraction: \[ 2 = 2x\sqrt{x} \]Divide by 2 and then square both sides: \[ 1 = x^3 \]Thus, \( x = 1 \) is a critical point in addition to \( x = 0 \).

Step 3: Test intervals around critical points

With critical points at \( x = 0 \) and \( x = 1 \), we choose test points in the intervals \((0, 1)\) and \((1, \infty)\):- For \( x \in (0,1) \) (e.g., \( x = 0.5 \)), \( g'(0.5) = \frac{2}{\sqrt{0.5}} - 1 \), which is positive, indicating increase.- For \( x \in (1, \infty) \) (e.g., \( x = 2 \)), \( g'(2) = \frac{2}{\sqrt{2}} - 4 \), which is negative, indicating decrease.

Step 4: Analyze increasing and decreasing intervals

From the sign analysis, the function is increasing on the interval \( (0, 1) \) and decreasing on the interval \( (1, \infty) \).

Step 5: Determine local and absolute extrema

Check the values of \( g(x) \) at critical points and intervals' endpoints:\( g(0) = 4\sqrt{0} - 0^2 + 3 = 3 \)\( g(1) = 4\sqrt{1} - 1^2 + 3 = 6 \)At \( x = 1 \), \( g(x) \) has a local maximum of 6. The function decreases in \((1, \infty)\), hence no other maximum point value exceeds 6 within the domain. The point \( x = 1 \) also presents the absolute maximum.

Key Concepts

Local Extrema

Local extrema refer to points in a function where it reaches a peak or a valley within a given neighborhood. In simpler terms, it's where the function has a local high point (local maximum) or local low point (local minimum). For the function given, which is \[ g(x) = 4\sqrt{x} - x^2 + 3 \], we first find the derivative to locate these potential turning points.

• We identified critical points by setting the derivative \( g'(x) = \frac{2}{\sqrt{x}} - 2x \) to zero.
• This process revealed that \( x = 1 \) is a critical point.
• We determined the behavior of the function around these points, checking whether the function is increasing or decreasing.

Around \( x = 1 \), the function goes from increasing to decreasing. This shift signifies a peak, indicating that \( x = 1 \) is a local maximum and can be graphically visualized as a hilltop.

Absolute Extrema

Absolute extrema are the highest and lowest values that a function reaches over its entire domain. These peaks and troughs are not limited to just a neighborhood but an entire domain. For the function \( g(x) = 4\sqrt{x} - x^2 + 3 \):

• By evaluating at the critical points and the limits of the domain, we determine the absolute extrema. Here, we compared \( g(0) = 3 \) and \( g(1) = 6 \).
• The function has an absolute maximum at \( x = 1 \) where \( g(x) = 6 \).
• Because the function decreases after \( x = 1 \), it does not reach any value greater than 6, thus confirming \( x = 1 \) is indeed the absolute maximum.
• Since the domain doesn't include negative or undefined points, no absolute minimum is discussed in this scenario.

Derivative

The derivative of a function helps us understand how the function's output changes with respect to changes in the input. For the function \( g(x) = 4\sqrt{x} - x^2 + 3 \), the derivative \( g'(x) \) is calculated as follows:\[ g'(x) = \frac{2}{\sqrt{x}} - 2x \]This derivative shows us how steep the curve is, telling us where the function increases or decreases.

• This information is crucial for finding intervals and characteristics like local and absolute extrema.
• When \( g'(x) = 0 \), it indicates potential extrema, while the sign (+ or -) shows increasing (positive) or decreasing (negative) behavior.
• By evaluating \( g'(x) \) across different intervals, we determine the behavior of the function between critical points, providing insights into overall function behavior.

Utilizing a derivative is akin to having a compass, guiding us through the highs and lows of a function, offering a clearer image of its landscape.