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Determine where the graph of the function is concave upward and where it is concave downward. Also, find all inflection points of the function. $$ f(x)=x-\sin x, \quad 0 \leq x \leq 4 \pi $$

Short Answer

Expert verified
The graph of the function \(f(x) = x - \sin x\) is concave upward in intervals (0 , \(\pi\)) and (2\(\pi\), 3\(\pi\)), concave downward in intervals (\(\pi\), 2\(\pi\)) and (3\(\pi\), 4\(\pi\)), and has inflection points at x = 0, \(\pi\), 2\(\pi\), 3\(\pi\), 4\(\pi\).

Step by step solution

01

Find the first derivative of f(x)

The given function is \(f(x) = x - \sin x\). To find its derivative, apply the differentiation rules: \(f'(x) = \frac{d}{dx}(x - \sin x) = 1 - \cos x\)
02

Find the second derivative of f(x)

Now, differentiate the first derivative with respect to x to find the second derivative: \(f''(x) = \frac{d}{dx}(1 - \cos x) = \sin x\)
03

Determine concavity

To find where the function is concave upward, find where the second derivative is greater than 0: \(f''(x) = \sin x > 0\) Sin is positive in the first and second quadrants. As \(0 \leq x \leq 4 \pi\), the function is concave upward in the intervals \((0 , \pi)\) and \(2\pi+x \in (2\pi, 3\pi)\) for any x in the first quadrant. To find where the function is concave downward, find where the second derivative is less than 0: \(f''(x) = \sin x < 0\) Sin is negative in the third and fourth quadrants. So, the function is concave downward in the intervals \(\pi+x \in (\pi, 2\pi)\) and \(3\pi+x \in (3\pi, 4\pi)\) for any x in the first quadrant.
04

Inflection points

To find the inflection points, set the second derivative equal to 0 and solve for x: \(f''(x) = \sin x = 0\) The solutions for sin x equal to 0 in the specified interval are \(x = 0, \pi, 2\pi, 3\pi, 4\pi\). These are the inflection points of the function. In conclusion, the graph of the function f(x) is: - Concave upward in intervals (0 , \(\pi\)) and (2\(\pi\), 3\(\pi\)) - Concave downward in intervals (\(\pi\), 2\(\pi\)) and (3\(\pi\), 4\(\pi\)) - The inflection points are at x = 0, \(\pi\), 2\(\pi\), 3\(\pi\), 4\(\pi\)

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Key Concepts

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

First Derivative
The first derivative of a function is a critical concept in calculus. It provides information about the rate at which the function changes. For any function \( f(x) \), the first derivative, denoted \( f'(x) \), reveals a lot about the function's behavior.
If \( f'(x) > 0 \) over a range of \( x \)-values, the function is increasing on that interval. If \( f'(x) < 0 \), the function is decreasing.
When it comes to our exercise, we have a function \( f(x) = x - \sin x \). The first derivative is calculated as follows:
  • Differentiate \( x \), which is \( 1 \).
  • Differentiate \(-\sin x\), which is \(-\cos x\).
Thus, the first derivative is:
\[ f'(x) = 1 - \cos x \]
This expression tells us how the slope of the tangent to the curve changes at each point along \( x \). Understanding this helps in determining the critical points of the function.
Second Derivative
The second derivative of a function gives us further insight into its concavity, telling us whether the graph of the function is curving upwards or downwards. The second derivative, \( f''(x) \), is simply the derivative of the first derivative.
For \( f(x) = x - \sin x \), with \( f'(x) = 1 - \cos x \), finding the second derivative involves differentiating again:
  • Differentiate the constant \( 1 \), resulting in \( 0 \).
  • Differentiate \(-\cos x\), resulting in \( \sin x \).
Thus, we have:
\[ f''(x) = \sin x \]
The second derivative clues us into concavity:
  • If \( f''(x) > 0 \), the function is concave upward (like a smile).
  • If \( f''(x) < 0 \), the function is concave downward (like a frown).
This is pivotal for identifying intervals of concavity and potential inflection points.
Inflection Point
An inflection point on the graph of a function is a point where the concavity changes, from concave up to concave down or vice versa. These points are found where the second derivative equals zero and changes sign.
For our function \( f(x) = x - \sin x \), we find the second derivative as \( f''(x) = \sin x \). Setting \( \sin x = 0 \) to find inflection points, we solve:
\(x = 0, \pi, 2\pi, 3\pi, 4\pi\) within the interval \([0, 4\pi]\).
These points are where the function switches concavity, and understanding them helps us sketch the function graph accurately.
The inflection points are essential indicators that help describe the function's geometry.
Trigonometric Functions
Trigonometric functions, such as \( \sin x \) and \( \cos x \), are fundamental in calculus due to their periodic nature and derivatives properties. These functions frequently appear in various mathematical contexts.
The sine function \( \sin x \):
  • Is periodic, with a period of \(2\pi\).
  • Takes values between \(-1\) and \(1\).
  • Is positive in the first and second quadrants.
  • Is zero at integer multiples of \(\pi\).
The cosine function \( \cos x \) is closely related, being the sine function shifted by \(\frac{\pi}{2}\). Understanding these properties helps in analyzing equations involving trigonometric functions and identifying points of interest, such as maximum, minimum, and inflection points.
These functions are vital tools in solving problems involving periodic behavior and wave patterns in physics and engineering.

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