Problem 56 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the area of the region bounded by the graph of the function \(f(x)=(x+1) / \sqrt{x}\), the \(x\) -axis, and the lines \(x=a\) and \(x=b\) where \(a\) and \(b\) are, respectively, the \(x\) -coordinates of the relative minimum point and the inflection point of \(f\)

Step-by-Step Solution

Answer

The area of the region bounded by the graph of the function \(f(x) = (x+1)/\sqrt{x}\), the x-axis, and the lines \(x=a\) and \(x=b\) where \(a\) and \(b\) are the x-coordinates of the relative minimum and inflection points, respectively, is \(\frac{35}{3}\) square units.

Step 1: Derive the given function

In order to find the values for a and b, we first need to derive the function. The derivative of the function \(f(x)=(x+1) / \sqrt{x}\) is computed using the quotient rule of differentiation, which is \((u'v-uv')/(v^2)\) where \(u = x+1\) and \(v = \sqrt{x}\). Hence, the derivative is given by \[f'(x) = \frac{(1)\sqrt{x}-(x+1)(1/2\sqrt{x})}{x}= \frac{2\sqrt{x}-x-1}{2x}\]

Step 2: Find the relative minimum point

A relative minimum point occurs when the derivative of the function changes sign from negative to positive. By setting the derivative equation equal to zero, the critical points are found. So, we get: \[ \frac{2\sqrt{x}-x-1}{2x} = 0\] Solving this equation reveals that the critical points are \(x=1\) and \(x=4\). The first derivative test will determine if each of these points is a minimum, maximum, or neither.

Step 3: Apply the first derivative test

The first derivative test checks for the sign of the derivative around the critical point. A change from negative to positive indicates a relative minimum. Checking around \(x=1\) and \(x=4\), it is observed that \(f'(x)\) changes from positive to negative at \(f'(1)\) indicating a relative maximum and from negative to positive at \(f'(4)\) indicating a relative minimum. Therefore, \(a=4\).

Step 4: Find the second derivative and the inflection point

An inflection point is where the second derivative of the function changes sign. To find this, compute the second derivative and set it to 0. The second derivative of \(f(x)\) is found by simply taking the derivative of \(f'(x)\). Setting this equal to zero will yield the points at which the function changes concavity. This gives \(x = b = 1\).

Step 5: Evaluate integral from a to b

The area under the curve from \(a\) to \(b\) is given by the definite integral of the function over this interval. In this case, the area will be: \[A=\int_{a}^{b} f(x) dx = \int_{1}^{4} (x+1)/\sqrt{x} dx \] Evaluating this definite integral will give the area of the region bounded by the graph of the function, the x-axis, and the lines \(x=a\) and \(x=b\). The definite integral equals \[\frac{14}{3} + 4\sqrt{4} - \frac{4}{3} -\sqrt{1}=\frac{20}{3}+8-1=\frac{35}{3}\] Therefore, the area of the region bounded by the graph of the function \(f(x) = (x+1)/\sqrt{x}\), the x-axis, and the lines \(x=a\) and \(x=b\) is \(\frac{35}{3}\) square units.

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. Think of it like the speed of a car; it's telling you how fast you're going at an exact moment. To find the derivative of the function \( f(x) = \frac{x+1}{\sqrt{x}} \), we use the quotient rule. The quotient rule helps us differentiate functions that are ratios: \( \frac{u}{v} \), where \( u \) and \( v \) are both functions of \( x \). The rule is given by \( (u'v - uv')/(v^2) \).

For our function, \( u = x + 1 \) and \( v = \sqrt{x} \). Calculating the derivative, we get:
\[ f'(x) = \frac{2\sqrt{x} - x - 1}{2x} \]

• \( u' = 1 \) (derivative of \( x+1 \))
• \( v' = \frac{1}{2\sqrt{x}} \) (derivative of \( \sqrt{x} \))
• Simplifying gives \( f'(x) = \frac{2\sqrt{x} - x - 1}{2x} \)

Derivatives help us find critical points like maxima and minima by setting \( f'(x) = 0 \) and solving.

Integral

An integral can be thought of as the area under a curve. If you imagine drawing a curve on the graph, the integral is like calculating how much paint you need to fill the space under that curve. To find the area between the curve \( f(x) = \frac{x+1}{\sqrt{x}} \) and the \( x \)-axis from the points \( x = 1 \) to \( x = 4 \), we calculate the definite integral:
\[ A = \int_{1}^{4} \frac{x+1}{\sqrt{x}} \, dx \]
The integration process adds up infinite tiny pieces under the curve. Once evaluated, this gives us the numerical area, which is \( \frac{35}{3} \) square units.

Critical Points

Critical points occur where the derivative equals zero or is undefined; these are potential maxima, minima, or saddle points. For our function, the critical points are identified by solving \( f'(x) = 0 \):

• The equation \( \frac{2\sqrt{x} - x - 1}{2x} = 0 \) helps find critical points at \( x = 1 \) and \( x = 4 \).
• A first derivative test tells us if these points are minima, maxima, or neither. Looking around \( x = 4 \), the sign change from negative to positive confirms a minimum.

Locating these points helps narrow down regions of interest in the graph, useful for integration.

Inflection Point

An inflection point is where the graph changes its shape, bending from concave up to concave down or vice versa. It occurs where the second derivative changes sign. For this function, taking the second derivative of the first derivative \( f'(x) \) leads to finding the inflection point.

• Compute \( f''(x) \) by differentiating \( f'(x) \).
• Setting \( f''(x) = 0 \) and solving will identify \( x = 1 \) as the inflection point.

Understanding inflection points gives insight into the concavity of functions and how steepness changes over an interval.