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NeTakaya
3 years ago
15

Here we will study the function f (x) = e ^ x sin (x), where x ∈ [0, 2π]. a) Determine where the function is decreasing and incr

easing. 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.

Mathematics
2 answers:
Semmy [17]3 years ago
3 0

fff

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}


Vilka [71]3 years ago
3 0

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}

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