Consider fx exsinx on 0 x 2pi Find Domain Intercepts Symm

Consider

f(x) = (e^-x))*(sin(x) on 0 < x < 2pi

Find

Domain:

Intercepts:

Symmetry:

Asymptodes:

Intervals of Increase and Decrease:

Local Max and Min:

Convavity and points of inflection:

Sketch the curve:

Please be thorough I am a beginner.

Solution

As the graph has been shown above, the answers to other questions are- Domain- 0 to 2pi as e^-x and sin(x) have complete domain y-intercept- put x=0, so y=0 x-intercept- as e^-x cannot be zero so sin(x)=0 for x=0,pi,2pi so x=0,pi,2pi are the x-intercept. as sin is an odd function, so the curve is symmetric about the origin. no asymptodes differentiate f(x). we get f\'(x)= (-e^-x)*cos(x) so cos(x)=0 give points of local max and min and points of inflection max=pi/2, min=3pi/2 interval of increase=(0,pi/2), (3pi/2,2pi) decrease=(pi/2,3pi/2) f\'\'(x) gives u the points of concavity