Find all points on the graph of the function 
f(x) = 2 sin(x) + sin2(x)
 at which the tangent line is horizontal. (Use n as your arbitrary integer.)
(x, y)
=	

    
Incorrect: Your answer is incorrect.
 

 (smaller y-value)
(x, y)
=	

    
Incorrect: Your answer is incorrect.
 

 (larger y-value)


To find the points on the graph of the function \( f(x) = 2\sin(x) + \sin^2(x) \) where the tangent line is horizontal, we need to determine where the derivative of the function \( f'(x) \) is equal to zero. A horizontal tangent line occurs where the slope is zero, which means the first derivative must equal zero.

### Step 1: Differentiate the function
The given function is:
\[
f(x) = 2\sin(x) + \sin^2(x)
\]
We will differentiate this with respect to \( x \).

The derivative of \( 2\sin(x) \) is:
\[
\frac{d}{dx}[2\sin(x)] = 2\cos(x)
\]

The derivative of \( \sin^2(x) \) (using the chain rule) is:
\[
\frac{d}{dx}[\sin^2(x)] = 2\sin(x)\cos(x)
\]

Thus, the derivative of \( f(x) \) is:
\[
f'(x) = 2\cos(x) + 2\sin(x)\cos(x)
\]

### Step 2: Set the derivative equal to zero
We need to find where the tangent is horizontal, i.e., where \( f'(x) = 0 \):
\[
2\cos(x) + 2\sin(x)\cos(x) = 0
\]

Factor out \( 2\cos(x) \):
\[
2\cos(x)(1 + \sin(x)) = 0
\]

This equation will be zero when either factor is zero.

#### Case 1: \( \cos(x) = 0 \)
\[
\cos(x) = 0 \implies x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}
\]

#### Case 2: \( 1 + \sin(x) = 0 \)
\[
1 + \sin(x) = 0 \implies \sin(x) = -1
\]
This occurs at:
\[
x = \frac{3\pi}{2} + 2n\pi, \quad n \in \mathbb{Z}
\]

### Step 3: Find corresponding \( y \)-values
Now that we have the \( x \)-coordinates, we find the corresponding \( y \)-values by substituting these values into the original function \( f(x) \).

#### Case 1: \( x = \frac{\pi}{2} + n\pi \)
For \( x = \frac{\pi}{2} \):
\[
f\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) + \sin^2\left(\frac{\pi}{2}\right) = 2(1) + 1^2 = 3
\]
Thus, the point is \( \left(\frac{\pi}{2}, 3\right) \).

For \( x = \frac{3\pi}{2} \):
\[
f\left(\frac{3\pi}{2}\right) = 2\sin\left(\frac{3\pi}{2}\right) + \sin^2\left(\frac{3\pi}{2}\right) = 2(-1) + (-1)^2 = -2 + 1 = -1
\]
Thus, the point is \( \left(\frac{3\pi}{2}, -1\right) \).

### Final Answer:
The points at which the tangent line is horizontal are:
\[
(x, y) = \left(\frac{\pi}{2}, 3\right)
\]
and
\[
(x, y) = \left(\frac{3\pi}{2}, -1\right)
\]

Would you like more details or have any questions? Here are 5 related questions you might explore:

1. What is the general formula for finding horizontal tangents in a function?
2. How would the procedure change if the function involved \( \cos^2(x) \) instead of \( \sin^2(x) \)?
3. What other types of critical points, besides horizontal tangents, can be found using derivatives?
4. How would the result change if the original function was \( f(x) = 2\sin(x) + \cos^2(x) \)?
5. How does the graph of \( f(x) = 2\sin(x) + \sin^2(x) \) compare to simpler sine and cosine functions?

**Tip**: Always remember to check both the first and second derivative to classify the type of critical points!

Find all points on the graph of the function f(x) = 2 sin(x) + sin^2(x) at which the tangent line is horizontal.

Explore how to find points where the tangent line is horizontal on the graph of f(x) = 2sin(x) + sin^2(x). This detailed solution involves key calculus concepts, including differentiation of trigonometric functions and critical points. Ideal for students in Grades 11-12 or college calculus.

Math Problem Statement

Find all points on the graph of the function f(x) = 2 sin(x) + sin2(x) at which the tangent line is horizontal. (Use n as your arbitrary integer.) (x, y)

(smaller y-value) (x, y)

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation