Respuesta :
we are given
[tex]f(x)=e^x sin(x)[/tex]
(a)
Firstly, we will find critical numbers
so, we will find derivative
[tex]f'(x)=e^x sin(x)+e^x cos(x)[/tex]
now, we can set it to 0
and then we can solve for x
we get
[tex]x=\frac{3\pi }{4} ,x=\frac{7\pi }{4}[/tex]
now, we can draw a number line and then locate these values
and then we can find sign of derivative on each intervals
increasing intervals:
[tex][0,\frac{3\pi}{4} )U(\frac{7\pi}{4} , 2\pi][/tex]
Decreasing interval:
[tex](\frac{3\pi}{4} ,\frac{7\pi}{4} )[/tex]
(b)
Local maxima:
It is the value of x where function changes from increasing to decreasing
so, local maxima is at
[tex]x=\frac{3\pi}{4}[/tex]
Local minima:
It is the value of x where function changes from decreasing to increasing
so, local minima is at
[tex]x=\frac{7\pi}{4}[/tex]
now, we will plug critical numbers and end values into original function
and we get
At x=0:
[tex]f(0)=e^0 sin(0)[/tex]
[tex]f(0)=0[/tex]
At [tex]x=\frac{3\pi}{4}[/tex]:
[tex]f(\frac{3\pi}{4})=e^{\frac{3\pi}{4}} sin(\frac{3\pi}{4})[/tex]
[tex]f(\frac{3\pi}{4})=7.46049[/tex]
At [tex]x=\frac{7\pi}{4}[/tex]:
[tex]f(\frac{7\pi}{4})=e^{\frac{7\pi}{4}} sin(\frac{7\pi}{4})[/tex]
[tex]f(\frac{7\pi}{4})=-172.640[/tex]
At [tex]x=2\pi [/tex]:
[tex]f(2\pi)=e^{2\pi} sin(2\pi )[/tex]
[tex]f(2\pi )=0[/tex]
Global maxima:
It is the largest value among them
so, we get
[tex]f(\frac{3\pi}{4})=7.46049[/tex]
Global minima:
It is the largest value among them
so, we get
[tex]f(\frac{7\pi}{4})=-172.640[/tex]
(c)
now, we can find second derivative
[tex]f'(x)=e^x sin(x)+e^x cos(x)[/tex]
[tex]f''(x)=\frac{d}{dx}\left(e^x\sin \left(x\right)+e^x\cos \left(x\right)\right)[/tex]
[tex]=\frac{d}{dx}\left(e^x\sin \left(x\right)\right)+\frac{d}{dx}\left(e^x\cos \left(x\right)\right)[/tex]
[tex]=e^x\sin \left(x\right)+\cos \left(x\right)e^x+e^x\cos \left(x\right)-e^x\sin \left(x\right)[/tex]
[tex]f''(x)=2e^x\cos \left(x\right)[/tex]
now, we can set it to 0
and then we can solve for x
[tex]f''(x)=2e^x\cos \left(x\right)=0[/tex]
so, we get
[tex]x=\frac{\pi}{2} ,x=\frac{3\pi}{2}[/tex]
now, we can draw number line and locate these values
and then we can find sign of second derivative on each intervals
concave up intervals:
[tex][0,\frac{\pi}{2})U(\frac{3\pi}{2}, 2\pi][/tex]
Concave down intervals:
[tex](\frac{\pi}{2} ,\frac{3\pi}{2})[/tex]
Turning points:
All values of x for which concavity changes
so, we get turning points at
[tex]x=\frac{\pi}{2} ,x=\frac{3\pi}{2}[/tex]
fff
[tex]f(x) = e^x sin (x)[/tex]
To find increasing and decreasing intervals we take derivative
[tex]f'(x) = e^xsin(x)+e^x(cosx)= e^x(sinx+cosx)[/tex]
Now we set the derivative =0 and solve for x
[tex]e^x(sinx+cosx)=0[/tex]
sinx + cosx =0
divide whole equation by cos x
[tex]\frac{sinx}{cosx} + \frac{cosx}{cosx} =0[/tex]
tanx +1 =0
tanx = 1
[tex]x=\frac{3\pi }{4}[/tex] and [tex]x=\frac{7\pi}{4}[/tex]
Now we pick a number between 0 to [tex]\frac{3\pi }{4}[/tex]
Lets pick [tex]\frac{\pi }{2}[/tex]
Plug it into the derivative
[tex]f'(x) =e^{\frac{\pi }{2}}(sin(\frac{\pi}{2})+cos(\frac{\pi }{2}))[/tex]
= 4.810 is positive
So the graph of f(x) is increasing on the interval [0, [tex]x=\frac{3\pi }{4}[/tex])
Now we pick a number between [tex]\frac{7\pi}{4}[/tex] to 2pi
Lets pick [tex]\frac{11\pi}{6}[/tex]
Plug it into the derivative
[tex]f'(x) =e^{\frac{11\pi}{6}}(sin(\frac{11\pi}{6})+cos(\frac{11\pi }{6}))[/tex]
= 116 is positive
So the graph of f(x) is increasing on the interval [tex](\frac{7\pi }{4}, 2\pi)[/tex]
Increasing interval is [tex](0,\frac{3\pi }{4}) U (\frac{7\pi }{4}, 2\pi)[/tex]
Decreasing interval is [tex](\frac{3\pi}{4}, \frac{7\pi}{4})[/tex]
(b)
The graph of f(x) increases and reaches a local maximum at [tex]x=\frac{3\pi}{4}[/tex]
The graph of f(x) decreases and reaches a local minimum at [tex]x=\frac{7\pi}{4}[/tex]
(c)
f(0) = 0
[tex]f(2\pi)=0[/tex]
[tex]f(\frac{3\pi }{4})=7.46[/tex]
[tex]f(\frac{7\pi}{4})=-172.64[/tex]
Here global maximum at [tex]x=\frac{3\pi}{4}[/tex]
Here global minimum at [tex]x=\frac{7\pi}{4}[/tex]
