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

smaller -value larger -value
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 Reed Morgan. in the United States

Mar 22,2025

Question

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

smaller -value larger -value
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 Reed Morgan. in the United States

Mar 22,2025

UpStudy AI · Accepted Answer

### Step 1. Find the Second Derivative The function given is $$ f(x)=x+9\cos(x). $$ First, we compute the first derivative: $$ f'(x)=1-9\sin(x). $$ Next, we compute the second derivative: $$ f''(x)=-9\cos(x). $$ ### Step 2. Find Points of Inflection Inflection points occur where the concavity changes. This is where the second derivative is zero and changes sign. Set the second derivative equal to zero: $$ -9\cos(x)=0 \quad \Longrightarrow \quad \cos(x)=0. $$ Within the interval $$[0,2\pi]$$, the solutions to $$\cos(x)=0$$ are: $$ x=\frac{\pi}{2} \quad \text{and} \quad x=\frac{3\pi}{2}. $$ Now, calculate the corresponding $$y$$-values: - For $$x=\frac{\pi}{2}$$: $$ f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}+9\cos\left(\frac{\pi}{2}\right)=\frac{\pi}{2}+9(0)=\frac{\pi}{2}. $$ So the point is $$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$$. - For $$x=\frac{3\pi}{2}$$: $$ f\left(\frac{3\pi}{2}\right)=\frac{3\pi}{2}+9\cos\left(\frac{3\pi}{2}\right)=\frac{3\pi}{2}+9(0)=\frac{3\pi}{2}. $$ So the point is $$\left(\frac{3\pi}{2}, \frac{3\pi}{2}\right)$$. ### Step 3. Determine Concavity Intervals The concavity of the graph is determined by the sign of $$ f''(x) $$: - **Concave Upward:** $$ f''(x)>0 $$ We have: $$ f''(x)=-9\cos(x)>0 \quad \Longrightarrow \quad \cos(x)<0. $$ This inequality is true when: $$ x\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right). $$ - **Concave Downward:** $$ f''(x)<0 $$ We have: $$ f''(x)=-9\cos(x)<0 \quad \Longrightarrow \quad \cos(x)>0. $$ This inequality is true when: $$ x\in\left(0,\frac{\pi}{2}\right) \quad \text{and} \quad x\in\left(\frac{3\pi}{2},2\pi\right). $$ ### Final Answers - **Inflection Points:** - Smaller $$ x $$-value: $$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$$ - Larger $$ x $$-value: $$\left(\frac{3\pi}{2}, \frac{3\pi}{2}\right)$$ - **Concavity:** - Concave upward on: $$\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$$ - Concave downward on: $$\left(0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$$ - **Inflection Points:** - Smaller $$ x $$-value: $$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$$ - Larger $$ x $$-value: $$\left(\frac{3\pi}{2}, \frac{3\pi}{2}\right)$$ - **Concavity:** - Concave upward on: $$\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$$ - Concave downward on: $$\left(0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$$

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Solución

Step 1. Find the Second Derivative

Step 2. Find Points of Inflection

Step 3. Determine Concavity Intervals

Final Answers

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