Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 29

y=e^xsin(x),-pi<=x<=pi
1) Asymptotes:
There is no undefined point, so the function has no vertical asymptote.
Since -oo and oo are not in the domain of the function , so it has no horizontal asymptote.
2) Intercepts:
plug in x=0 in the function, to get y intercept,
y=e^0sin(0)=0
3) Maxima/Minima/Critical Points
Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.
y'=e^xcos(x)+sin(x)e^x
y'=e^x(cos(x)+sin(x))
Now to find the critical numbers, solve for x for y'=0,
e^x(cos(x)+sin(x))=0
e^x=0 has no solution
cos(x)+sin(x)=0
sin(x)=-cos(x)
tan(x)=-1
Solutions for the range are
x=-pi/4 , x=(3pi)/4
plug in the values of x in the function to get the y-coordinate
f(-pi/4)=e^(-pi/4)sin(-pi/4)=e^(-pi/4)*(-1/sqrt(2))=(-e^(-pi/4))/sqrt(2)
f((3pi)/4)=e^((3pi)/4)sin((3pi)/4)=e^((3pi)/4)/sqrt(2)
Absolute Maximum=e^((3pi)/4)/sqrt(2) at x=(3pi)/4
Absolute Minimum=(-e^(-pi/4))/sqrt(2) at x=-pi/4
Intervals of Increase/Decrease
Let's check the signs of y' by plugging test points in the intervals (-pi,-pi/4),(-pi/4,(3pi)/4) ,((3pi)/4,pi) ,
f'(-pi/2)=e^(-pi/2)(sin(-pi/2)+cos(-pi/2))=e^(-pi/2)(-1+0)=-e^(-pi/2)
f'(0)=e^0(sin(0)+cos(0))=1(0+1)=1
f'((7pi)/8)=e^((7pi)/8)(sin((7pi)/8)+cos((7pi)/8))=e^((7pi/8))(0.38268+(-0.92388))
f'((7pi)/8)=-0.5412e^((7pi)/8)
Since f'(-pi/2),f'((7pi)/8) are negative, so the function is decreasing in the intervals (-pi,-pi/4) , ((3pi)/4,pi)
Since f'(0) is positive , so the function is increasing in the interval (-pi/4,(3pi)/4)
4) Inflection points and Concavity:
y'=e^x(sinx+cos(x))
y''=e^x(cos(x)-sin(x))+(sin(x)+cos(x))e^x
y''=e^x(cos(x)-sin(x)+sin(x)+cos(x))
y''=2cos(x)e^x
set y''=0 and solve for x,
2cos(x)e^x=0
e^x=0 has no solution,
For cos(x)=0
solutions for the range -pi<=x<=pi are,
x=-pi/2, x=pi/2
Now let's check the sign of y'' by plugging test values in the intervals (-pi,-pi/2),(-pi/2,pi/2),(pi/2,pi)
f''(-(3pi)/4)=2e^((-3pi)/4)cos((-3pi)/4)=2e^(-3pi/4)*(-1/sqrt(2))=-sqrt(2)e^((-3pi)/4)
f''(0)=2e^0cos(0)=2
f''((3pi)/4)=2e^((3pi)/4)cos((3pi)/4)=2e^((3pi)/4)*(-1/sqrt(2))=-sqrt(2)e^((3pi)/4)
Since f''((-3pi)/4),f''((3pi)/4) are negative, so the function is concave down in the intervals (-pi,-pi/2),(pi/2,pi)
Since f''(0) is positive , so the function is concave up in the interval (-pi/2,pi/2)
Function sketched is attached.