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, and asymptotes, and find the intervals on which the function is increasing or decreasing, and the intervals on which the function is concave up or concave down. $$f(x)=\frac{\sin \pi x}{1+\sin \pi x} \text { on }[0,2]$$

Short answer

The x-intercepts are (0, 0) and (1, 0); the y-intercept is (0, 0). The critical points occur at $$x = \frac{1}{4}$$ and $$x = \frac{3}{4}$$. The inflection point is at $$x = \frac{1}{2}$$. The function is decreasing on intervals $$[0, \frac{1}{4}]$$ and $$[1, \frac{3}{4}]$$, increasing on the interval $$[\frac{1}{4}, 1]$$, and constant on the interval $$[\frac{3}{4}, 2]$$. The function is concave up on the interval $$[0, \frac{1}{2}]$$, concave down on the interval $$[\frac{1}{2}, 1]$$, and constant on the interval $$[1, 2]$$.

Step-by-step solution

Intercepts

To find the x-intercepts, set $$f(x) = 0$$: $$0 = \frac{\sin \pi x}{1+\sin \pi x}$$ This equation simplifies to: $$\sin \pi x = 0$$ The x-intercepts occur when $$x = 0$$ and $$x = 1$$, so the x-intercepts are (0, 0) and (1, 0). For the y-intercept, we evaluate $$f(0) = \frac{\sin(0)}{1+\sin(0)} = 0$$. So the y-intercept is (0, 0).

Local Extrema

To find the local extrema, we need to find the critical points by setting the first derivative equal to 0: $$f'(x) = \frac{\pi (1+\sin \pi x)(\cos \pi x) - \pi \cos \pi x(\sin \pi x)}{(1+\sin \pi x)^2}$$ This simplifies to: $$f'(x)=\frac{-\pi\sin(\pi x)\cos(\pi x)}{(1+\sin\pi x)^2}$$ Setting $$f'(x) = 0$$, we find that $$x = \frac{1}{4}$$ and $$x = \frac{3}{4}$$ are critical points.

Inflection Points

To find the inflection points, we need to find where the second derivative changes sign: $$f''(x) = \frac{\pi^2\sin(\pi x)\left[(1+\sin(\pi x))^2-2\sin(\pi x)\cos^2(\pi x)\right]}{(1+\sin(\pi x))^4}$$ Setting $$f''(x) = 0$$, we find that $$x = \frac{1}{2}$$ is an inflection point.

Asymptotes

This function doesn't have any vertical, horizontal, or oblique asymptotes.

Increasing or Decreasing Intervals

By analyzing the first derivative, we find that - The function is decreasing on intervals $$[0, \frac{1}{4}]$$ and $$[1, \frac{3}{4}]$$ - The function is increasing on the interval $$[\frac{1}{4}, 1]$$ - The function is constant on the interval $$[\frac{3}{4}, 2]$$

Concave Up or Concave Down Intervals

By analyzing the second derivative, we find that - The function is concave up on the interval $$[0, \frac{1}{2}]$$ - The function is concave down on the interval $$[\frac{1}{2}, 1]$$ - The function is constant on the interval $$[1, 2]$$ Now, we know all the characteristics of the function. We can use a graphing utility to create the complete graph of the function and label the intercepts, local extrema, inflection points, and intervals on the graph.

Key concepts

X-Intercepts

When we begin graphing functions, identifying the x-intercepts is a crucial first step. These are the points where the function crosses the x-axis, which occurs when the output value, or y, is zero. For the function f(x) = \(\frac{\sin \pi x}{1+\sin \pi x}\) on the interval [0,2], the x-intercepts are found by setting f(x) equal to zero and solving for x.

This resulted in the x-intercepts being (0,0) and (1,0), indicating that at these points, the curve touches or crosses the x-axis. Knowing where these intercepts are helps establish a foundational understanding of the function's behavior.

Y-Intercepts

Similarly to x-intercepts, the y-intercept of a function is where the curve crosses the y-axis. This is found by evaluating the function at x=0. For our function, evaluating f(0) showed that the y-intercept also happens to be at the origin (0,0).

This singular point of intersection provides an important reference as it's one of the few definitive points we can use to start sketching the curve. Additionally, since the y-intercept and one of the x-intercepts coincide, it emphasizes the need to consider the entire behavior of the function, not just individual points.

Local Extrema

Functions often have high and low points, known as local extrema. These are the peaks (maximums) and valleys (minimums) where the function's graph changes direction from increasing to decreasing or vice versa. To locate the local extrema of a function, we set its first derivative to zero, because at these points the slope of the tangent to the curve is horizontal.

Using calculus methods, we've determined that the function has critical points at x = 1/4 and x = 3/4. These points are candidates for local extrema and are essential for understanding the function's overall landscape.

Inflection Points

Moving on to inflection points, these are points on the curve where the function changes concavity, from concave up to concave down or vice versa. They are found by setting the second derivative of the function to zero or identifying where it does not exist. These points signify a change in the curvature of the graph.

For this function, there is an inflection point at x = 1/2. It's where the concavity switches, giving us further detail about the geometry of our function's graph.

Asymptotes

While many functions have asymptotes—lines that the graph approaches but never reaches—our function here lack such features. Recognizing whether the function has vertical, horizontal, or oblique asymptotes is crucial for graphing because they outline the boundaries of the graph and indicate behavior at the extremities of the domain. In this case, we simply note the absence of asymptotes as part of the function's characteristics.

Intervals of Increase and Decrease

Understanding where a function is increasing or decreasing helps in sketching the general shape of the graph. By analyzing the first derivative, we determine the direction of the function's slope. If the first derivative is positive over an interval, the function is increasing there. Conversely, if it's negative, the function is decreasing.

Our function is decreasing on the intervals [0, 1/4] and [1, 3/4], and increasing on [1/4, 1]. Between [3/4, 2], the function's derivative is zero, implying that the function's value is constant.

Concavity of Functions

The concavity of a function tells us about the curvature of the graph. If a function is concave up, the graph looks like a valley with the opening upward, and if it's concave down, it looks like a mountain peak. This is determined by the sign of the second derivative: a positive second derivative indicates the function is concave up, and a negative one indicates concave down.

For the function in question, our analysis revealed it is concave up on [0, 1/2] and concave down on [1/2, 1]. After x=1, the second derivative suggests no further change in concavity, which aligns with the derivative being zero, resulting in a constant function.