#### To determine

**To discuss:** the graph y=e−xsinx;0≤x≤2π

#### Answer

Please see the graph above

#### Explanation

A. Domain:(0,2π)

B. Intercepts: Put x=0, we get y=0 as the y intercept. Put y=0, we get x=0,π,2π as the x intercept.

C. Symmetry: The curve is not symmetrical about the X-axis and Y-axis

D. Asymptotes: NO vertcal asymptote and horizontal asymptote

E. Interval of Increasing or decreasing: The function y=e−xsinx ;0≤x≤2π is increasing in (0,π4) and (3π4,2π) and decreasing in (π4,3π4)

F. Local maxima or minima : We have y=e−xsinx; 0≤x≤2π

Differentiating with respect to x

y′=e−xcosx+sinx(−e−x)=e−x(cosx−sinx)

y′=0⇒e−x(cosx−sinx)=0⇒cosx−sinx=0⇒tanx=1⇒x=π4,5π4

The local maxima occurs at x=π4

y(π4)=e−π4sinπ4=0.3224

The local minima occurs at x=5π4

y(5π4)=e−5π4sin5π4=−0.0139

G. **Inflection Points :** For inflection points y″=0 and the curve is concave up if y″>0 and concave down if y″<0

y′=e−x(cosx−sinx)y″=e−x(−cosx−sinx)−e−x(cosx−sinx)=−2e−xcosx

**Inflection points are** (π2,3π2)

**The curve is concave up in** (π2,3π2) and concave down in (0,π2) and (3π2,2π)

H. Please See the graph below

**Conclusion:** Please see the graph above