Question

Use a graphing utility together with analytical methods to create a complete graph of the following functions. Be sure to find and label the intercepts, local extrema, inflection points, asymptotes, intervals where the function is increasing/decreasing, and intervals of concavity. $$f(x)=\frac{x \sin x}{x^{2}+1} \text { on }[-2 \pi, 2 \pi]$$

Short answer

Question: Graph the function $$f(x) = \frac{x\sin x}{x^2+1}$$ on the interval $$[-2\pi, 2\pi]$$ by determining the intercepts, local extrema, inflection points, asymptotes, intervals where the function is increasing/decreasing, and intervals of concavity. Solution: 1. There are no x-intercepts. The y-intercept is (0,0). 2. The first derivative is $$f'(x) = \frac{(x^2+1)(\cos x)+x\sin x(-2x)}{(x^2+1)^2}$$. The second derivative is $$f''(x) = \frac{((x^2+1)(-2x\sin x)+(\cos x)(2x))(x^2+1)^2-(-2x\sin x+2\cos x)(x^2+1)(2(x^2+1))}{(x^2+1)^4}$$. 3. The critical points occur at approximately $$x = -1.895, 0, 1.895$$. The possible points of inflection occur at approximately $$x = -2.09, 0, 2.09$$. 4. The function is increasing on the intervals $$(-2\pi, -1.895)$$ and $$(1.895, 2\pi)$$, and decreasing on the intervals $$(-1.895, 0)$$ and $$(0, 1.895)$$. There are local maxima at approximately $$x = -1.895$$ and $$x = 1.895$$, and a local minimum at $$x = 0$$. 5. The function is concave up on the intervals $$(-2\pi, -2.09)$$ and $$(2.09, 2\pi)$$, and concave down on the interval $$(-2.09, 0)$$ and $$(0, 2.09)$$. There are inflection points at approximately $$x = -2.09, 0, $$ and $$2.09$$. By plotting these points and analyzing the intervals of increasing/decreasing and concavity, you can draw the graph of the function on the interval $$[-2\pi, 2\pi]$$ using these characteristics.

Step-by-step solution

Find the x and y intercepts

To find the x intercepts, set $$f(x)$$ to zero and solve for x: $$0=\frac{x\sin x}{x^2+1}$$ To find the y intercepts, set x to zero and find $$f(0)$$: $$f(0)=\frac{0\cdot \sin 0}{0^2+1}$$

Calculate the first and second derivatives

Now, we'll find the first and second derivatives of the function using the Quotient Rule: First derivative, $$f'(x)$$: $$f'(x)=\frac{(x^2+1)(\cos x)+x\sin x(-2x)}{(x^2+1)^2}$$ Second derivative, $$f''(x)$$: $$f''(x)=\frac{((x^2+1)(-2x\sin x)+(\cos x)(2x))(x^2+1)^2-(-2x\sin x+2\cos x)(x^2+1)(2(x^2+1))}{(x^2+1)^4}$$

Determine the critical points and possible points of inflection

Find when the first derivative is equal to zero or does not exist: $$0=\frac{(x^2+1)(\cos x)+x\sin x(-2x)}{(x^2+1)^2}$$ Solve this equation to find the critical points. Find when the second derivative is equal to zero or does not exist: $$0=\frac{((x^2+1)(-2x\sin x)+(\cos x)(2x))(x^2+1)^2-(-2x\sin x+2\cos x)(x^2+1)(2(x^2+1))}{(x^2+1)^4}$$ Solve this equation to find the possible points of inflection.

Analyze intervals of increase/decrease and local extrema

Using the critical points found in Step 3, determine the intervals where $$f'(x)>0$$ (increasing) and $$f'(x)<0$$ (decreasing). Check if a change in the sign of $$f'(x)$$ occurs at these critical points. If there's a sign change, it indicates a local extrema.

Analyze concavity and inflection points

Using the possible points of inflection found in Step 3, determine the intervals where $$f''(x)>0$$ (concave up) and $$f''(x)<0$$ (concave down). If there's a sign change in $$f''(x)$$ at a point, then it is an inflection point.

Identify any asymptotes

Since the function is a rational function, there are no vertical asymptotes. For horizontal asymptotes, analyze the limits as x approaches $$\pm \infty$$. In this case, it's not applicable since the domain is restricted to $$[-2\pi, 2\pi]$$.

Combine all the information to graph the function

Using all the information obtained from steps 1 through 6 (intercepts, local extrema, inflection points, asymptotes, intervals of increasing/decreasing, and intervals of concavity), plot all the points and draw the graph of the function on the interval $$[-2\pi, 2\pi]$$. Additionally, you can use a graphing utility to verify your work.

Key concepts

X and Y Intercepts

In graphing rational functions, intercepts are points where the graph crosses the axes. To find the x-intercepts, we set the numerator to zero and solve for x, as these are the points where the function's value is zero. For the function \( f(x)=\frac{x \sin x}{x^{2}+1} \), there are potentially multiple x-intercepts due to the \(\sin x\) term, which we evaluate within the given interval \( [-2\pi, 2\pi] \).

For the y-intercept, we set x to zero and evaluate the function, which gives us a single point on the y-axis. In this case, \( f(0)=\frac{0\cdot \sin 0}{0^2+1} = 0 \), so the y-intercept is at the origin, (0,0). Recognizing where the function crosses the axes is crucial for understanding the overall graph.

First and Second Derivatives

Derivatives provide us with information about the slope of the graph at any given point. The first derivative, \( f'(x) \), indicates the gradient of the tangent line to the curve at a particular x-value, which helps us determine whether the function is increasing or decreasing at that point.

The second derivative, \( f''(x) \), offers insights into the concavity of the graph—whether it is curving up or down. For the function \( f(x)=\frac{x \sin x}{x^{2}+1} \), we use the Quotient Rule to calculate these derivatives. Having both derivatives is essential for finding critical points, inflection points, and understanding the function's concavity.

Critical Points

Critical points occur where the first derivative \( f'(x) \) is either zero or undefined. These points are important as they could potentially be local maximums, local minimums, or saddle points—places where the function's graph changes direction. To find them, we set the first derivative equal to zero and solve for x. Additionally, if the derivative doesn't exist at some point, that could also indicate a critical point, though this isn't the case with our example function.

Inflection Points

Inflection points are where the function's concavity changes, signified by the second derivative \( f''(x) \) changing its sign. To determine inflection points, we solve for when \( f''(x) \) equals zero or does not exist. These points are valuable for understanding the behavior of a graph, as they indicate where the curve changes from being 'concave up' to 'concave down' or vice versa.

Intervals of Increase/Decrease

By analyzing the first derivative, we can identify intervals where the function is increasing (\( f'(x)>0 \)) or decreasing (\( f'(x)<0 \)). After finding the critical points, we test intervals around those points to determine where the function's slope is positive or negative. These intervals are pivotal for sketching the function since they show where the graph rises and falls.

Concavity

The concept of concavity deals with the direction of the curve's bend. If the second derivative is positive (\( f''(x)>0 \)), the graph is concave up, resembling a 'U' shape, while a negative value (\( f''(x)<0 \)) indicates the graph is concave down, like an inverted 'U'. Understanding concavity helps us to predict the shape of the graph between inflection points and is essential for accurate graph drawing.

Asymptotes

Asymptotes are lines that the graph approaches but never touches. Vertical asymptotes occur at values of x where the function goes to infinity, often where the denominator of a rational function is zero. However, since our function has the denominator \( x^2+1 \) which is never zero, there are no vertical asymptotes. Horizontal asymptotes are found by examining the end behavior as x approaches infinity—but in this example, the domain is limited to \( [-2\pi, 2\pi] \), so horizontal asymptotes are not applicable. Even though our function doesn't have asymptotes, in general, they are crucial for understanding the long-term behavior of rational functions.