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In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ g(x)=\frac{\sin x}{1+\sin x} $$

Short Answer

Expert verified
The function g(x) = \(\frac{\sin{x}}{1+\sin{x}}\) has: 1. Domain: \((-\infty, \infty) \) and range: \([-0.5, 0]\) 2. x-intercepts at \(x = n\pi\), y-intercept at (0, 0) 3. No asymptotes 4. No symmetry 5. Critical points (when \(\cos{x} = 0\)) 6. Concavity change with the second derivative By combining these properties, you can sketch the graph of the given function.

Step by step solution

01

Domain and Range

The function is a rational function with sine in the numerator and an addition of sine plus one in the denominator. Since sine is defined for all real numbers, there is no restriction on the domain of \(x\). Domain: \((-\infty, \infty)\) Now, let's find the range. The sine function takes values from -1 to 1. Minimum value of the denominator will be \(1 + (-1) = 0\), which means it will never be zero and the function will always be defined. If \(x=0\), \(\sin x = 0\) and g(x) = 0. The maximum value \(\sin{x}\) can take is 1. Therefore, the maximum value g(x) can take is \(\frac{1}{1 + 1} = 0.5\). Thus, the range is \([-0.5, 0]\).
02

Intercepts

x-intercepts (zeros) occur when \(g(x) = 0\), which happens when \(\sin x = 0\). Since these zeros occur at integer multiples of \pi, we have x-intercepts at \(x = n\pi\), where \(n\) is an integer: \(x = \dots, -2\pi, -\pi, 0, \pi, 2\pi, \dots\) y-intercept occurs when \(x = 0\): \(g(0) = 0\). So the y-intercept is the point (0, 0).
03

Asymptotes

No vertical asymptotes can occur, as the denominator would have to become 0, which cannot happen since \(1 + \sin x\) is never 0. Now, let's analyze the limits at the infinity: \[ \lim_{x \to -\infty} \frac{\sin x}{1 + \sin x} = \lim_{x \to \infty} \frac{\sin x}{1 + \sin x} \] Since \(\sin{x}\) oscillates between -1 and 1, this limit does not exist. Hence, there is no horizontal asymptote.
04

Symmetry

Since \(\sin(-x) = -\sin(x)\), \[ g(-x) = \frac{\sin(-x)}{1 + \sin(-x)} = \frac{-\sin x}{1 - \sin x} \] g(-x) is not equal to g(x) nor -g(x), so our function is neither even nor odd and has no symmetry.
05

Derivative and Critical Points

Now, let's find the first derivative to extract information about critical points and where the function is increasing and decreasing. \(g'(x) = \frac{d}{dx} \left(\frac{\sin{x}}{1+\sin{x}}\right)\) Applying the Quotient Rule, we get: \(g'(x) = \frac{(1+\sin{x})\cos{x} - \sin{x}\cos{x}}{(1+\sin{x})^2}\) Simplifying the expression, we obtain: \(g'(x) = \frac{\cos{x}}{(1+\sin{x})^2}\) g'(x) is 0 when \(x = \frac{2n+1}{2}\pi\) and undefined when \(x = (2n+1)\pi\), where \(n\) is an integer. We only care about the zeros of the first derivative, as they indicate the critical points of the function.
06

Second Derivative and Points of Inflection

Now, we will find the second derivative to analyze the concavity of the function. \(g''(x) = \frac{d^2}{dx^2} \left(\frac{\sin{x}}{1+\sin{x}}\right)\) Applying the Quotient Rule again, we get: \(g''(x) = \frac{\{(1+\sin{x})^2*(-\sin{x})-(\cos{x})^2(1+\sin{x})(2)\}}{(1+\sin{x})^4}\) Simplifying the expression, we obtain: \(g''(x) = -\frac{(1+\sin{x})(\sin{x} + 2\cos^2{x})}{(1+\sin{x})^3}\)
07

Finally Sketching the Graph

Now, we need to combine all of the information gathered above to sketch the graph. 1. Domain: \((-\infty, \infty) \) and range: \([-0.5, 0]\) 2. Intercepts: x-intercepts at \(x = n\pi\), y-intercept at (0, 0) 3. No asymptotes 4. No symmetry 5. Critical points (when \(\cos{x} = 0\)) 6. Concavity change with the second derivative By combining these properties, you will be able to sketch the graph of the given function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain and Range of a Function
Understanding the domain and range of a function is like knowing the limits within which the function can operate. The domain refers to all the possible input values, while the range is about the potential outputs.

For the rational function in our exercise, \( g(x) = \frac{\sin x}{1 + \sin x} \), we see there are no restrictions on the domain because the sine function accepts all real numbers, hence, the domain is \((-\infty, \infty)\).

Now, let’s talk about the range. Since the minimum value of the denominator is 1 (as sine swings from -1 to 1), the function never touches zero, thus avoiding any potential vertical asymptotes. We can determine the range by considering the boundaries set by the sine function; the lowest output value for \(g(x)\) is when \(\sin x\) is -1, leading up to a maximum of 0.5 when \(\sin x\) is 1.

For our function, the range expertly navigates between these limits, giving us a range of \([-0.5, 0]\).
Graphing Rational Functions
Graphing a rational function like \(g(x)\) in the given exercise involves a blend of calculus and pre-calculus concepts.

First up, we look at intercepts. Our x-intercepts are points where \(g(x) = 0\), which, thanks to our friendly sine function, happens at multiples of \(\pi\). And a special shoutout to the origin (0,0) for being the lone y-intercept!

Then, we move to asymptotes. But wait, for \(g(x)\), there's no vertical or horizontal asymptote to worry about since the denominator is always non-zero, and the values of sine swing back and forth without settling, as we approach infinity.

Finally, we’ve got to consider symmetry—our function doesn’t seem to favor the mirror images at all. We’ve now laid out a roadmap that allows us to calmly approach the task of actually plotting this graph on paper – or screen!
Trigonometric Functions
Let's swing into the sinusoidal world of trigonometric functions. They're like the ocean waves of mathematics - sometimes high, sometimes low, but always in motion.

In our function, \(\sin x\) is the star of the show. It oscillates between -1 and 1, and these values are crucial for the behaviour of our rational function. The periodic nature of sine—with its peaks and valleys at regular intervals—plays a massive part in the graph's appearance: those x-intercepts at multiples of \(\pi\) are a direct result of this trig function's rhythmic dance.

Moreover, the sine function's waviness ensures that our rational function varies smoothly, which makes sketching it an artful process. But even more wonderful? This sine wave's characteristics feed into the calculus concepts we’ll need for the rest of our analysis. Isn't it amazing how trigonometry and calculus can join hands to paint a beautiful picture of \(g(x)\)?
First and Second Derivatives
Now, put on your calculus cap because it's time to talk derivatives – the mathematical equivalent of a detective looking for clues on a graph's behaviour. The first derivative, \(g'(x)\), tells us where the function is increasing or decreasing and reveals those mysterious critical points where the slope is flat as a pancake.

For \(g(x)\), we get the first derivative using the quotient rule and find that it equals \(\frac{\cos{x}}{(1+\sin{x})^2}\). Critical points come into view when the cosine reaches zero, cueing us on where to find potential hills and valleys on the graph.

But wait, there's more! The second derivative, \(g''(x)\), is all about the concavity—think of it as the graph's curvy posture. By calculating it, also with the quotient rule, we’re checking for inflection points, where the graph switches from a convex (smiling) to concave (frowning) shape, or vice versa.

Together, these two derivatives are the dynamic duo for understanding the up-and-down, and back-and-forth nature of our function's graph. They're essential for curve sketching, so keep these tools handy whenever you’re facing a rational function with a trigonometric twist!

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