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Chapter 3: Problem 16

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=2 \csc \frac{3 x}{2}, \quad(0,2 \pi)\)

Short Answer

Expert verified
After taking the second derivative, setting it equal to zero, and finding discontinuities, the function's concavity changes at the inflection points. These points, as well as the intervals of concavity, are found by testing the second derivative on both sides of these points.

Step by step solution

01

Differentiate the function.

To find the points of inflection, we first have to find the second derivative of the function. But before that, let's find the first derivative. \The derivative of csc(u), in general, is \(-\csc(u) \cot(u) \cdot u'\), where \(u'\) is the derivative of \(u\). \For \(f(x) = 2 \csc( \frac{3x}{2})\), let \(u = \frac{3x}{2}\) and \(u' = \frac{3}{2}\). So, \ \(f'(x) = -2\[\csc(\frac{3x}{2}) \cot(\frac{3x}{2})\] * \frac{3}{2}\
02

Compute the second derivative.

Again using trigonometry rules, since the derivative of \(\csc(u)\) is \(-\csc(u)\cot(u)\) and of \(\cot(u)\) is \(-\csc^2(u)\), the second derivative will be:\(f''(x) = \[\frac{3}{2} \csc(\frac{3x}{2}) \cot(\frac{3x}{2})\][2 \csc(\frac{3x}{2}) \cot(\frac{3x}{2}) - \frac{3}{2} \csc^2(\frac{3x}{2})]\)
03

Find the points where the second derivative equals 0 find undefined points

This is done by solving the equation \(f''(x) = 0\) and find points where the derivative is undefined. For the function \(f''(x)\) to equal zero, the function inside the brackets has to equal zero, because \(\csc(x)\) is never zero. So, place \(2 \csc(\frac{3x}{2}) \cot(\frac{3x}{2}) - \frac{3}{2} \csc^2(\frac{3x}{2}) = 0\) and solve for \(x\).
04

Analyze the concavity and inflection points

Use the results from Step 3 to find the points of inflection, where the graph changes concavity, and the intervals where the graph is concave up or concave down. By substituting the values of \(x\) found in step 3 into \(f''(x)\) and analyzing whether \(f''(x)\) has different signs on either side of the point. Create a sign chart for the concavity and note that if it changes from + to - , the graph is concave down and if it changes from - to +, the function is concave up.

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Most popular questions from this chapter

A section of highway connecting two hillsides with grades of \(6 \%\) and \(4 \%\) is to be built between two points that are separated by a horizontal distance of 2000 feet (see figure). At the point where the two hillsides come together, there is a 50 -foot difference in elevation. (a) Design a section of highway connecting the hillsides modeled by the function \(f(x)=a x^{3}+b x^{2}+c x+d\) \((-1000 \leq x \leq 1000)\). At the points \(A\) and \(B,\) the slope of the model must match the grade of the hillside. (b) Use a graphing utility to graph the model. (c) Use a graphing utility to graph the derivative of the model. (d) Determine the grade at the steepest part of the transitional section of the highway.

In Exercises \(75-86\), use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptotes that exist. $$ f(x)=\frac{x}{x^{2}-4} $$

In Exercises \(75-86\), use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptotes that exist. $$ f(x)=2+\left(x^{2}-3\right) e^{-x} $$

A line with slope \(m\) passes through the point (0,-2) . (a) Write the distance \(d\) between the line and the point (4,2) as a function of \(m\). (b) Use a graphing utility to graph the equation in part (a). (c) Find \(\lim _{m \rightarrow \infty} d(m)\) and \(\lim _{m \rightarrow-\infty} d(m)\). Interpret the results geometrically.

In Exercises \(75-86\), use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptotes that exist. $$ f(x)=\frac{2 \sin 2 x}{x} $$
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