- #1
ttpp1124
- 110
- 4
- Homework Statement
- can someone confirm my work?
- Relevant Equations
- n/a
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).
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.
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.
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.
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.