Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

Determine the open intervals on which the graph of the function is concave upward or concave downward. (Enter your answers
using interval notation. If an answer does not exist, enter DNE.)
concave upward
concave downward

Ask by Pollard Hill. in the United States

Mar 21,2025

Question

Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

Determine the open intervals on which the graph of the function is concave upward or concave downward. (Enter your answers
using interval notation. If an answer does not exist, enter DNE.)
concave upward
concave downward

Ask by Pollard Hill. in the United States

Mar 21,2025

UpStudy AI · Accepted Answer

We begin with the given function $$ f(x)=10\csc\left(\frac{3x}{2}\right). $$ Let $$ \theta=\frac{3x}{2}. $$ Then the function is written as $$ f(x)=10\csc\theta. $$ **Step 1. Compute the First Derivative** Recall that $$ \frac{d}{d\theta}\left(\csc\theta\right)=-\csc\theta\cot\theta. $$ By the chain rule, $$ f'(x)=10\cdot\frac{d}{dx}\left(\csc\theta\right)=10\cdot\left(-\csc\theta\cot\theta\right)\cdot\frac{d\theta}{dx}. $$ Since $$ \frac{d\theta}{dx}=\frac{3}{2}, $$ we have $$ f'(x)=-10\csc\theta\cot\theta\cdot\frac{3}{2}=-15\csc\theta\cot\theta. $$ In terms of $$ x $$, $$ f'(x)=-15\csc\left(\frac{3x}{2}\right)\cot\left(\frac{3x}{2}\right). $$ **Step 2. Compute the Second Derivative** Differentiate $$ f'(x)=-15\csc\theta\cot\theta $$ with respect to $$ x $$: $$ f''(x)=-15\cdot\frac{d}{dx}\left[\csc\theta\cot\theta\right]. $$ Again, using the chain rule the differentiation with respect to $$ x $$ can be written as: $$ \frac{d}{dx}\left[\csc\theta\cot\theta\right]=\frac{d}{d\theta}\left[\csc\theta\cot\theta\right]\cdot\frac{d\theta}{dx}. $$ We already know that $$ \frac{d\theta}{dx}=\frac{3}{2}. $$ Now, differentiate $$ h(\theta)=\csc\theta\cot\theta $$ with respect to $$ \theta $$. We use the product rule: $$ h'(\theta)=\left(\csc\theta\right)'\cot\theta+\csc\theta\left(\cot\theta\right)'. $$ Recall that $$ \left(\csc\theta\right)'=-\csc\theta\cot\theta \quad\text{and}\quad \left(\cot\theta\right)'=-\csc^2\theta. $$ Thus, $$ h'(\theta)=-\csc\theta\cot\theta\cdot\cot\theta+\csc\theta\cdot\left(-\csc^2\theta\right)=-\csc\theta\cot^2\theta-\csc^3\theta. $$ Now, substituting back, we have: $$ \frac{d}{dx}\left[\csc\theta\cot\theta\right]=\left(-\csc\theta\cot^2\theta-\csc^3\theta\right)\cdot\frac{3}{2}. $$ Then, $$ f''(x)=-15\cdot\left(-\csc\theta\cot^2\theta-\csc^3\theta\right)\cdot\frac{3}{2}. $$ Simplify the negatives: $$ f''(x)=15\cdot\frac{3}{2}\left(\csc\theta\cot^2\theta+\csc^3\theta\right)=\frac{45}{2}\left(\csc\theta\cot^2\theta+\csc^3\theta\right). $$ Expressing in terms of $$ x $$, $$ f''(x)=\frac{45}{2}\left[\csc\left(\frac{3x}{2}\right)\cot^2\left(\frac{3x}{2}\right)+\csc^3\left(\frac{3x}{2}\right)\right]. $$ **Step 3. Finding the Point of Inflection** Inflection points occur where $$ f''(x) $$ is zero or changes sign **and** $$ f(x) $$ is continuous at that point. Notice that in the expression for $$ f''(x) $$, $$ f''(x)=\frac{45}{2}\csc\left(\frac{3x}{2}\right)\left[\cot^2\left(\frac{3x}{2}\right)+\csc^2\left(\frac{3x}{2}\right)\right], $$ the term $$ \cot^2\left(\frac{3x}{2}\right)+\csc^2\left(\frac{3x}{2}\right) $$ is always positive (since both squared functions are always nonnegative and $$\csc^2\theta \geq 1$$ when defined). Thus, $$ f''(x) $$ can only be zero if $$ \csc\left(\frac{3x}{2}\right)=0. $$ But $$\csc\theta=\frac{1}{\sin\theta}$$ is never zero because $$ \frac{1}{\sin\theta}=0 $$ would require $$ \sin\theta $$ to be infinite, which is impossible. Therefore, there is no $$ x $$ in $$(0, 2\pi)$$ for which $$ f''(x)=0 $$ and the function $$ f(x) $$ usually has vertical asymptotes where the sine is 0. The suspected candidates where a sign change might occur are points where $$ f''(x) $$ is undefined. However, these are exactly the points where $$ f(x) $$ is not defined (i.e. where $$\sin\left(\frac{3x}{2}\right)=0$$). In the interval $$(0,2\pi)$$, these occur when $$ \frac{3x}{2}=n\pi \quad \text{or} \quad x=\frac{2n\pi}{3} $$ with $$ n $$ an integer. The values that lie in $$(0,2\pi)$$ are: $$ x=\frac{2\pi}{3} \quad \text{and} \quad x=\frac{4\pi}{3}. $$ Since these points are not in the domain of $$ f(x) $$, the function does not have any inflection points on $$(0,2\pi)$$. Thus, the point of inflection is: $$ (x, y)=(\mathrm{DNE}) $$. **Step 4. Determine the Open Intervals of Concavity** The sign of $$ f''(x) $$ is determined solely by the sign of $$ \csc\left(\frac{3x}{2}\right)=\frac{1}{\sin\left(\frac{3x}{2}\right)}. $$ Since the remaining factor, $$ \cot^2\left(\frac{3x}{2}\right)+\csc^2\left(\frac{3x}{2}\right) $$, is always positive, we have: - $$ f''(x)>0 $$ (concave upward) when $$ \csc\left(\frac{3x}{2}\right)>0 $$ which happens when $$ \sin\left(\frac{3x}{2}\right)>0, $$ - $$ f''(x)<0 $$ (concave downward) when $$ \sin\left(\frac{3x}{2}\right)<0 $$. Solve for each: 1. Find where $$ \sin\left(\frac{3x}{2}\right)>0 $$: - The sine function is positive when its argument is in the interval $$ (0, \pi) $$ modulo $$ 2\pi $$. - For the principal interval, $$ 0 < \frac{3x}{2} < \pi $$ implies $$ 0 < x < \frac{2\pi}{3}. $$ - For the next interval in $$(0,2\pi)$$, consider $$ 2\pi < \frac{3x}{2} < 3\pi $$. Dividing by $$ \frac{3}{2} $$ gives: $$ \frac{4\pi}{3} < x < 2\pi. $$ Therefore, $$ f(x) $$ is concave upward on $$ (0,\,\frac{2\pi}{3})\cup\left(\frac{4\pi}{3},\,2\pi\right). $$ 2. Find where $$ \sin\left(\frac{3x}{2}\right)<0 $$: - Sine is negative when its argument is in $$ (\pi, 2\pi) $$ modulo $$ 2\pi $$. - For $$ \pi < \frac{3x}{2} < 2\pi $$, solve: $$ \frac{2\pi}{3}< x < \frac{4\pi}{3}. $$ Thus, $$ f(x) $$ is concave downward on $$ \left(\frac{2\pi}{3},\,\frac{4\pi}{3}\right). $$ **Final Answers:** - Point of inflection: $$ (\mathrm{DNE}) $$ - Concave upward on: $$ (0,\,\frac{2\pi}{3})\cup\left(\frac{4\pi}{3},\,2\pi\right) $$ - Concave downward on: There is no point of inflection on the interval $$(0, 2\pi)$$. The function is concave upward on $$(0, \frac{2\pi}{3})$$ and $$(\frac{4\pi}{3}, 2\pi)$$, and concave downward on $$(\frac{2\pi}{3}, \frac{4\pi}{3})$$.

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