Problem 34 - University Calculus: Early Transcendentals

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. $$g(x)=x^{2} \sqrt{5-x}$$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, 4)\) and decreasing on \((4, 5)\). It has a local maximum of 16 at \(x=4\).

Step 1: Find the Domain

The function is defined as \( g(x) = x^2 \sqrt{5-x} \). For \( \sqrt{5-x} \) to be real and defined, we require \( 5-x \geq 0 \). Thus, \( x \leq 5 \). Also, since we're dealing with a square root, \( x \) must be within the domain where it is real, which is \( (-\infty, 5] \).

Step 2: Compute the Derivative

To determine where the function is increasing or decreasing, calculate the derivative \( g'(x) \). Using product and chain rules: Let \( u = x^2 \) and \( v = \sqrt{5-x} \), then \( g'(x) = u'v + uv'\).\( u' = 2x \), \( v = (5-x)^{1/2} \), and \( v' = \frac{-1}{2(5-x)^{1/2}} \).Therefore, \( g'(x) = 2x\sqrt{5-x} + x^2\frac{-1}{2\sqrt{5-x}} \).

Step 3: Simplify the Derivative

Simplify \( g'(x) = 2x\sqrt{5-x} - \frac{x^2}{2\sqrt{5-x}} \). To combine these, use a common denominator:\( g'(x) = \frac{4x(5-x) - x^2}{2\sqrt{5-x}} \)\( = \frac{4x(5-x) - x^2}{2\sqrt{5-x}} \)\( = \frac{20x - 4x^2 - x^2}{2\sqrt{5-x}} \)\( = \frac{20x - 5x^2}{2\sqrt{5-x}} \).

Step 4: Find Critical Points

Set \( g'(x) = 0 \) to find critical points. Solve:\( \frac{20x - 5x^2}{2\sqrt{5-x}} = 0 \rightarrow 20x - 5x^2 = 0 \)\( \Rightarrow x(20 - 5x) = 0 \) which implies \( x = 0 \) or \( x = 4 \). These are our critical points.

Step 5: Determine Intervals of Increase and Decrease

Evaluate \( g'(x) \) on intervals defined by critical points and endpoints. Consider intervals \((-\infty, 0)\), \((0, 4)\), and \((4, 5)\).- For \((-\infty, 0)\), choose test point \(x = -1\): \(g'(x) > 0 \) so \(g(x)\) is increasing.- For \((0, 4)\), choose test point \(x = 2\): \(g'(x) > 0 \) so \(g(x)\) is increasing.- For \((4, 5)\), choose test point \(x = 4.5\): \(g'(x) < 0 \) so \(g(x)\) is decreasing.Thus, \(g(x)\) is increasing on \((-\infty, 4)\) and decreasing on \((4, 5)\).

Step 6: Evaluate Extreme Values

Check function values at critical points and endpoints: \(x = 0, 4, 5\).- \(g(0) = 0^2\sqrt{5-0} = 0\).- \(g(4) = 4^2\sqrt{5-4} = 16\) is a local maximum.- \(g(5) = 5^2\sqrt{5-5} = 0\).There is a local maximum at \(x=4\) with value \(16\) and potentially at global points, as both endpoints have value \(0\).

Key Concepts

Derivative

In calculus, the derivative of a function is a key concept used to analyze how the function changes. The derivative, often denoted as \( f'(x) \), represents the slope of the tangent line to the curve at any point \( x \). For the function \( g(x) = x^2 \sqrt{5-x} \), finding the derivative helps us identify where the function is increasing or decreasing. To compute the derivative \( g'(x) \), we apply the product rule and the chain rule because \( g(x) \) is composed of two functions multiplied together: \( x^2 \) and \( \sqrt{5-x} \). This requires:

• Product Rule: \( (uv)' = u'v + uv' \), where \( u = x^2 \) and \( v = \sqrt{5-x} \).
• Chain Rule: Helps differentiate the composite function \( \sqrt{5-x} \).

The resulting derivative provides a formula we can simplify, aiding in further analysis of the function's behavior.

Critical Points

Critical points of a function are where its derivative is zero or undefined. These points are significant because they may indicate potential local maxima or minima. To find the critical points for \( g(x) = x^2 \sqrt{5-x} \), we solve the equation \( g'(x) = 0 \). This involves solving the simplified form of the derivative:\[ g'(x) = \frac{20x - 5x^2}{2\sqrt{5-x}} = 0.\]By factoring out \( x \), we find the critical points: \( x = 0 \) and \( x = 4 \). These values are crucial because they distinguish the intervals on which \( g(x) \) increases or decreases.

Increasing and Decreasing Intervals

Understanding whether a function is increasing or decreasing on specific intervals involves examining the sign of its derivative within those intervals. For \( g(x) = x^2 \sqrt{5-x} \), the derivative \( g'(x) \) must be evaluated around the critical points and endpoints. This provides:

• Increasing intervals indicate \( g'(x) > 0 \).
• Decreasing intervals indicate \( g'(x) < 0 \).

By checking values in these segments:

• \((-\infty, 0)\) and \((0, 4)\) are intervals where the function is increasing.
• \((4, 5)\) is where the function is decreasing.

These intervals help us understand where the function generally climbs or descends.

Local and Absolute Extremes

Local extremes are the highest or lowest points in a specific region of a function's graph. Absolute extremes, on the other hand, are the highest or lowest values over the entire domain. For \( g(x) = x^2 \sqrt{5-x} \):

• A local maximum occurs where \( g(x) \) switches from increasing to decreasing at \( x = 4 \), yielding a value of 16.
• Analyzing the endpoints \( x = 0 \) and \( x = 5 \), both yield a value of 0, suggesting they may be absolute extremes as \( g(x) \) neither exceeds nor dips below this elsewhere within its domain.

Identifying these points provides insight into the function's overall shape and behavior.