Differentiate and simplify: f(x) = (sinx)(e^-2x)

In summary, the derivative of f(x) is given by f'(x) = (cosx)(e^-2x) - 2(sin x)(e^-2x). To simplify f'(x), you can factor out the common term of e^-2x to get f'(x) = (e^-2x)(cosx - 2sinx). The critical points of f(x) can be found by setting f'(x) = 0 and solving for x, which results in x = π/6 and x = 5π/6. Since f(x) is a product of two trigonometric functions, it does not have a maximum or minimum value, instead it oscillates between -1 and
  • #1
ttpp1124
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Homework Statement
can someone confirm my work?
Relevant Equations
n/a
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  • #2
It's okay, but not as simple as it might be.
 

1. What is the derivative of f(x)?

The derivative of f(x) is given by f'(x) = (cosx)(e^-2x) + (-2sinx)(e^-2x), which can be simplified to f'(x) = (e^-2x)(cosx - 2sinx).

2. How do you find the critical points of f(x)?

To find the critical points of f(x), set the derivative f'(x) equal to 0 and solve for x. In this case, we get x = pi/4 or x = 3pi/4 as the critical points.

3. What is the concavity of the graph of f(x)?

The concavity of the graph of f(x) can be determined by looking at the second derivative f''(x). If f''(x) is positive, the graph is concave up. If f''(x) is negative, the graph is concave down. In this case, f''(x) = (e^-2x)(-2sinx - 4cosx), so the graph is concave down for x < pi/4 and x > 3pi/4, and concave up for pi/4 < x < 3pi/4.

4. What is the limit of f(x) as x approaches infinity?

The limit of f(x) as x approaches infinity can be found by looking at the leading term of the function, which in this case is e^-2x. As x approaches infinity, e^-2x approaches 0, so the limit is 0.

5. How do you graph f(x)?

To graph f(x), first plot the critical points and any other important points, such as x-intercepts or points where the function is undefined. Then, use the concavity and limit information to determine the shape of the graph between these points. Finally, sketch the graph using these points and the overall behavior of the function.

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