Problem 6 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ f(x)=\frac{x^{2}+3}{x-1} $$

Step-by-Step Solution

Answer

Function is increasing on \[(-\infty, -1)\], decreasing on \[(-1, 3)\], increasing on \[(3, \infty)\]. Concave up on \[(-\infty, 1)\], concave down on \[(1, \infty)\].

Step 1: Find the first derivative

To determine intervals of increase and decrease, first find the derivative of the function. For the given function \[f(x)=\frac{x^{2}+3}{x-1}\], use the quotient rule: \[(\frac{u}{v})' = \frac{u'v - uv'}{v^2}\]. Let \[u = x^2 + 3\] and \[v = x - 1\]. Then \[u' = 2x\] and \[v' = 1\]. So the first derivative is \[f'(x)=\frac{(2x)(x-1)-(x^{2}+3)(1)}{(x-1)^{2}}\].

Step 2: Simplify the first derivative

Simplify the first derivative found in Step 1: \[f'(x) = \frac{2x(x-1) - (x^2 + 3)}{(x-1)^2} = \frac{2x^2 - 2x - x^2 - 3}{(x-1)^2} = \frac{x^2 - 2x - 3}{(x-1)^2}\]. So \[f'(x) = \frac{(x-3)(x+1)}{(x-1)^2}\].

Step 3: Determine intervals of increase and decrease

Set \[f'(x) = 0\] and solve for \[x\]. So \[\frac{(x-3)(x+1)}{(x-1)^2} = 0\] gives \[x = 3\] or \[x = -1\]. The critical points are \[x = -1, 3\]. Check the sign of \[f'(x)\] on the intervals \[(-\infty, -1), (-1, 3), (3, \infty)\].

Step 4: Analyze concavity using the second derivative

Find the second derivative to determine concavity. First, \[f''(x)\] is obtained by differentiating \[f'(x) = \frac{(x-3)(x+1)}{(x-1)^2}\] using the quotient rule again. Simplify to find \[f''(x)\]. Then determine the sign of \[f''(x)\] at critical points.

Step 5: Determine intercepts and asymptotes

To find x-intercepts, set \[f(x) = 0\]. Solving \[\frac{x^2+3}{x-1} = 0\], there are no real roots since \[x^2+3 eq 0\]. To find the y-intercept, set \[x = 0\]: \[f(0) = \frac{3}{-1} = -3\]. Vertical asymptote is at \[x = 1\]. For horizontal asymptote, compare degrees of numerator and denominator. As x approaches infinity, \[f(x)\] behaves like \[x\], so there's no horizontal asymptote, but an oblique one.

Step 6: Identify key features of the graph

Plot the critical points, vertical asymptote (x = 1), intercepts: y-intercept at (0,−3). Note intervals from steps 3 and 4. Check concavity near inflection points. Sketch the function incorporating all analyzed features.

Key Concepts

first derivative

To understand the behavior of a function, we first need to find its first derivative. The first derivative of a function, denoted as \(f'(x)\), gives us valuable information about where the function is increasing or decreasing. For our given function \(f(x) = \frac{x^2 + 3}{x - 1}\), we use the quotient rule to find the first derivative. The quotient rule formula is: \(\frac{u}{v}' = \frac{u'v - uv'}{v^2}\). Here, let \(u = x^2 + 3\) and \(v = x - 1\). After applying the rule, we get: \(f'(x) = \frac{(2x)(x-1) - (x^2 + 3)(1)}{(x - 1)^2}\). Simplifying this, we get: \(f'(x) = \frac{(x-3)(x+1)}{(x-1)^2}\). This simplified form is crucial for analyzing the function’s behavior.

second derivative

The second derivative, denoted as \(f''(x)\), provides insight into the concavity of the function. Concavity tells us whether the function curves upwards or downwards at given intervals. To find \(f''(x)\), we differentiate the first derivative \(f'(x) = \frac{(x-3)(x+1)}{(x-1)^2}\) using the quotient rule again. The calculation becomes more complex, but the result will significantly aid in understanding how the function bends. Once simplified, the second derivative helps identify points of inflection—where the concavity changes.

intervals of increase and decrease

To determine where our function is increasing or decreasing, we analyze the sign of the first derivative \(f'(x) = \frac{(x-3)(x+1)}{(x-1)^2}\). Set \(f'(x) = 0\) to find critical points: \(x = 3\) or \(x = -1\). Next, check the sign of \(f'(x)\) around these critical points. For intervals:

• For \((- \frac{\text{∞}}, -1)\), \(f'(x)\) is positive, indicating the function is increasing.
• For \((-1, 3)\), \(f'(x)\) is negative, meaning the function is decreasing.
• For \((3, \frac{\text{∞}})\), \(f'(x)\) is positive, so the function is increasing again.

This analysis shows us how the function behaves over different intervals.

concavity

The concavity of a function explains how it curves, either up or down. By analyzing the second derivative \(f''(x)\), we can determine concave up or concave down intervals.

• If \(f''(x) > 0\), the function is concave up (shaped like a cup).
• If \(f''(x) < 0\), the function is concave down (shaped like a cap).

Points where \(f''(x) = 0\) or changes signs are inflection points, indicating a change in concavity. Identifying these points helps create an accurate graph.

graphing functions

Creating the graph combines all the previous steps. Start by marking critical points and intercepts, such as \(x = -1, 3\) and the y-intercept at \(f(0) = -3\). Also include asymptotes:

• Vertical asymptote at \(x = 1\).
• No horizontal asymptote due to the function's behavior as \(x\to \frac{\text{∞}}\).

Consider intervals of increase/decrease and concavity to sketch the curve accurately. Putting everything together, you can visualize the complete behavior of \(f(x)\). This approach ensures your graph accurately reflects the function’s properties.