Problem 30 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

The first derivative of a certain function is \(f^{\prime}(x)=x^{2}(5-x)\). a. On what intervals is \(f\) increasing? Decreasing? b. On what intervals is the graph of \(f\) concave up? Concave down? c. Find the \(x\) coordinates of the relative extrema and inflection points of \(f\). d. Sketch a possible graph of \(f(x)\).

Step-by-Step Solution

Answer

Function is increasing on \((-∞, 0)\) and \((0, 5)\), decreasing on \((5, ∞)\). Concave up on \((0, \frac{10}{3})\), concave down on \((-∞, 0)\) and \((\frac{10}{3}, ∞)\). Relative extremum at \(x = 5\) (max), and \(x = 0\) (min); inflection points at \(x = 0\) and \(x = \frac{10}{3}\).

Step 1: Determine Critical Points

To determine the intervals where the function is increasing or decreasing, first find the critical points of the function by setting the first derivative equal to zero and solving for x: \[ f^{\raisebox{.2em}{\tiny\(\boldsymbol{\top}\)}}(x) = x^{2}(5-x) = 0 \] Solving this equation gives the critical points: \[ x = 0 \] and \[ x = 5 \].

Step 2: Determine Increasing and Decreasing Intervals

Use the sign of the first derivative to determine where the function is increasing or decreasing. Perform a sign test around the critical points: For \( x eq [0,5] \), choose test points in each interval: - For \( x < 0 \) (e.g., \( x = -1 \)), \( f^{\raisebox{.2em}{\tiny\(\boldsymbol{\top}\)}}(-1) = (-1)^{2}(5 - (-1)) = 1 \times 6 = 6 \) (positive). - For \( 0 < x < 5 \) (e.g., \( x = 2 \)), \( f^{\raisebox{.2em}{\tiny\(\boldsymbol{\top}\)}}(2) = (2)^{2}(5-2) = 4 \times 3 = 12 \) (positive). - For \( x > 5 \) (e.g., \( x = 6 \)), \( f^{\raisebox{.2em}{\tiny\(\boldsymbol{\top}\)}}(6) = (6)^{2}(5-6) = 36 \times (-1) = -36 \) (negative). Therefore, the function \( f \) is increasing on the intervals \( (-\boldsymbol{\top}, 0)\) and \( (0, 5) \), and decreasing on the interval \( (5, \boldsymbol{\top}) \).

Step 3: Second Derivative for Concavity

To determine concavity, find the second derivative of the function: \[ f''(x) = \frac{d}{dx}(x^{2}(5-x)) = 2x(5-x) + x^{2}(-1) = 2x(5-x) - x^{2} = 10x - 3x^{2} \] Set the second derivative equal to zero to find possible inflection points: \[ 10x - 3x^{2} = 0 \] Solving this gives: \[ x(10 - 3x) = 0 \] Thus, \( x = 0 \) and \( x = \frac{10}{3} \).

Step 4: Determine Concavity Intervals

Use the sign of the second derivative to determine the intervals of concavity. Perform a sign test around the points \( x = 0 \) and \( x = \frac{10}{3} \): For \( x < 0 \) (e.g., \( x = -1 \)), \( f''(-1) = 10(-1) - 3(-1)^{2} = -10 - 3 = -13 \) (negative). - For \( 0 < x < \frac{10}{3} \) (e.g., \( x = 1 \)), \( f''(1) = 10(1) - 3(1)^{2} = 10 - 3 = 7 \) (positive). - For \( x > \frac{10}{3} \) (e.g., \( x = 4 \)), \( f''(4) = 10(4) - 3(4)^{2} = 40 - 48 = -8 \) (positive). Therefore, the function is concave down on \( (-\boldsymbol{\top}, 0) \), concave up on \( (0, \frac{10}{3}) \), and concave down again on \( (\frac{10}{3}, \boldsymbol{\top}) \).

Step 5: Identify Relative Extrema and Inflection Points

Relative extrema occur at the critical points. Since the function changes from increasing to decreasing at \( x = 5 \), there is a relative maximum at \( x = 5 \). The function changes from decreasing to increasing at \( x = 0 \), indicating a relative minimum. Inflection points occur where the concavity changes sign: at \( x = 0 \) and at \( x = \frac{10}{3} \).

Step 6: Sketch the Graph

Using the information gathered: intervals of increasing/decreasing, concavity, and the relative extrema and inflection points, sketch a rough graph. The function has: - A relative minimum at \( x = 0 \). - A relative maximum at \( x = 5 \). - Inflection points at \( x = 0 \) and \( x = \frac{10}{3} \).

Key Concepts

first derivative

The first derivative of a function, denoted as \( f'(x) \), helps us determine where the function is increasing or decreasing. By calculating \( f'(x) \), you find the slope of the function at any point, giving insight into its behavior. To find where the function changes direction, set the first derivative equal to zero and solve for \( x \). These solutions are your critical points. For example, given \( f'(x) = x^2(5-x) = 0 \), solving this gives \( x = 0 \) and \( x = 5 \). These are the critical points where the function might change from increasing to decreasing or vice versa.

second derivative

The second derivative of a function, denoted as \( f''(x) \), provides information on the concavity of the function's graph. To find the second derivative, differentiate \( f'(x) \) once again. For instance, if \( f'(x) = x^2(5-x) \), then \( f''(x) = 10x - 3x^2 \). Setting \( f''(x) \) equal to zero, \( 10x - 3x^2 = 0 \), and solving for \( x \) gives potential inflection points, locations where the concavity might change. Solving this example provides the points \( x = 0 \) and \( x = \frac{10}{3} \).

increasing and decreasing intervals

To find where a function is increasing or decreasing, perform a sign test around the critical points found from the first derivative. Choose test points in each interval divided by the critical points. If \( f'(x) \) is positive, the function is increasing; if \( f'(x) \) is negative, the function is decreasing. For example, with critical points \( x = 0 \) and \( x = 5 \), you might test values \( x = -1 \), \( x = 2 \), and \( x = 6 \). If the first derivative result is positive for \( -1 \) and \( 2 \), and negative for \( 6 \), then the function increases on \( (-\raisebox{.2em}{\tiny\(\boldsymbol{\top}\)}, 0) \) and \( (0, 5) \), and decreases on \( (5, \raisebox{.2em}{\tiny\(\boldsymbol{\top}\)}) \).

concavity

Concavity refers to the direction the curve of the function opens. A function is concave up when it opens upwards like a U, and concave down when it opens downwards like an inverted U. By checking the sign of the second derivative, you can determine the concavity. If \( f''(x) > 0 \), the function is concave up; if \( f''(x) < 0 \), it is concave down. Performing a sign test around the potential inflection points helps. For example, for \( x < 0 \), \( f''(-1) = -13 \) shows concave down, for \( 0 < x < \frac{10}{3} \), \( f''(1) = 7 \) shows concave up, and \( x > \frac{10}{3} \), \( f''(4) = -8 \) back to concave down.

inflection points

Inflection points are where a function changes concavity. They occur where the second derivative is zero or undefined and the concavity changes. To find these points, first solve \( f''(x) = 0 \). For example, \( 10x - 3x^2 = 0 \) provides \( x = 0 \) and \( x = \frac{10}{3} \). Verify a change in concavity around these points using a sign test for the second derivative. Inflection points signal where the graph of the function shifts from concave up to concave down, or vice versa.

relative extrema

Relative extrema are peaks (maximums) or valleys (minimums) in a function's graph. Relative maxima occur where the function changes from increasing to decreasing, and relative minima where it changes from decreasing to increasing. To find these points, use the critical points found from the first derivative. Co-relate these with the sign changes analyzed in increasing and decreasing intervals. For example, at \( x = 5 \), the function changes from increasing to decreasing indicating a relative maximum. At \( x = 0 \), it changes from decreasing to increasing, indicating a relative minimum. Combining these insights helps in sketching the graph, identifying both relative extrema and inflection points.