Question

Here we will study the function f (x) = e ^ x sin (x), where x ∈ [0, 2π]. a) Determine where the function is decreasing and increasing. b) Find all local maximam and minimam. Does the absolute (global) maximam / minimam have? c) Determine where f (x) curves up and down. Also find any turning points.

Answer 1

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$f(x) = e^x sin (x)$

To find increasing and decreasing intervals we take derivative

$f'(x) = e^xsin(x)+e^x(cosx)= e^x(sinx+cosx)$

Now we set the derivative =0  and solve for x

$e^x(sinx+cosx)=0$

sinx + cosx =0

divide whole equation by cos x

$\frac{sinx}{cosx} + \frac{cosx}{cosx} =0$

tanx +1 =0

tanx = 1

$x=\frac{3\pi }{4}$ and  $x=\frac{7\pi}{4}$

Now we pick a number between 0 to  $\frac{3\pi }{4}$

Lets pick  $\frac{\pi }{2}$

Plug it into the derivative

$f'(x) =e^{\frac{\pi }{2}}(sin(\frac{\pi}{2})+cos(\frac{\pi }{2}))$

= 4.810 is positive

So the graph of f(x) is increasing on the interval [0, $x=\frac{3\pi }{4}$)

Now we pick a number between   $\frac{7\pi}{4}$ to 2pi

Lets pick  $\frac{11\pi}{6}$

Plug it into the derivative

$f'(x) =e^{\frac{11\pi}{6}}(sin(\frac{11\pi}{6})+cos(\frac{11\pi }{6}))$

= 116 is positive

So the graph of f(x) is increasing on the interval $(\frac{7\pi }{4}, 2\pi)$

Increasing interval is $(0,\frac{3\pi }{4}) U (\frac{7\pi }{4}, 2\pi)$

Decreasing interval is $(\frac{3\pi}{4}, \frac{7\pi}{4})$

(b)

The graph of f(x) increases and reaches a local maximum at $x=\frac{3\pi}{4}$

The graph of f(x) decreases and reaches a local minimum at $x=\frac{7\pi}{4}$

(c)

f(0) = 0

$f(2\pi)=0$

$f(\frac{3\pi }{4})=7.46$

$f(\frac{7\pi}{4})=-172.64$

Here global maximum at $x=\frac{3\pi}{4}$

Here global minimum at $x=\frac{7\pi}{4}$

Answer 2

we are given

$f(x)=e^x sin(x)$

(a)

Firstly, we will find critical numbers

so, we will find derivative

$f'(x)=e^x sin(x)+e^x cos(x)$

now, we can set it to 0

and then we can solve for x

we get

$x=\frac{3\pi }{4} ,x=\frac{7\pi }{4}$

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:

$[0,\frac{3\pi}{4} )U(\frac{7\pi}{4} , 2\pi]$

Decreasing interval:

$(\frac{3\pi}{4} ,\frac{7\pi}{4} )$

(b)

Local maxima:

It is the value of x where function changes from increasing to decreasing

so, local maxima is at

$x=\frac{3\pi}{4}$

Local minima:

It is the value of x where function changes from decreasing to increasing

so, local minima is at

$x=\frac{7\pi}{4}$

now, we will plug critical numbers and end values into original function

and we get

At x=0:

$f(0)=e^0 sin(0)$

$f(0)=0$

At $x=\frac{3\pi}{4}$:

$f(\frac{3\pi}{4})=e^{\frac{3\pi}{4}} sin(\frac{3\pi}{4})$

$f(\frac{3\pi}{4})=7.46049$

At $x=\frac{7\pi}{4}$:

$f(\frac{7\pi}{4})=e^{\frac{7\pi}{4}} sin(\frac{7\pi}{4})$

$f(\frac{7\pi}{4})=-172.640$

At $x=2\pi$:

$f(2\pi)=e^{2\pi} sin(2\pi )$

$f(2\pi )=0$

Global maxima:

It is the largest value among them

so, we get

$f(\frac{3\pi}{4})=7.46049$

Global minima:

It is the largest value among them

so, we get

$f(\frac{7\pi}{4})=-172.640$

(c)

now, we can find second derivative

$f'(x)=e^x sin(x)+e^x cos(x)$

$f''(x)=\frac{d}{dx}\left(e^x\sin \left(x\right)+e^x\cos \left(x\right)\right)$

$=\frac{d}{dx}\left(e^x\sin \left(x\right)\right)+\frac{d}{dx}\left(e^x\cos \left(x\right)\right)$

$=e^x\sin \left(x\right)+\cos \left(x\right)e^x+e^x\cos \left(x\right)-e^x\sin \left(x\right)$

$f''(x)=2e^x\cos \left(x\right)$

now, we can set it to 0

and then we can solve for x

$f''(x)=2e^x\cos \left(x\right)=0$

so, we get

$x=\frac{\pi}{2} ,x=\frac{3\pi}{2}$

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:

$[0,\frac{\pi}{2})U(\frac{3\pi}{2}, 2\pi]$

Concave down intervals:

$(\frac{\pi}{2} ,\frac{3\pi}{2})$

Turning points:

All values of x for which concavity changes

so, we get turning points at

$x=\frac{\pi}{2} ,x=\frac{3\pi}{2}$