Question

Need some help with the analytic study and graphic representation of the following function:

Answer 1

\[f:f(x)=(x-2)e ^{3x-1}\]

Answer 2

What exactly do you need?

Answer 3

First I want to know the limits of that funtion as x approaches + and - inifinity.
I got that both gave me infinity, but I believe I did something wrong there:

Answer 4

How did you find your limits?

Answer 5

\[\lim_{x \rightarrow +\infty}(x-2)e ^{3x-1}=+\infty \]

Answer 6

what made you conclude that?

Answer 7

Happens to be that you're right. Just want to hear your argument.

Answer 8

Well, I know that any number multiplied by infinity is equal to inifnity, and if I substract -1 it'll still be infinity, so I would end un havinf e^inf wich is equal to inifnity.
(x-2), if I replace infinity there, I would end up having infinity, so the total would be:
(infinity)(infinity) = infinity, and since I'm doing it from the right... +infinity,

Answer 9

agree.

What about as x------> - infinity?

Answer 10

That's where my problem starts:
So if I replace infinity it would be 0?

Answer 11

It does approach 0.

Answer 12

as x approaches negative infinity, e^(3x-1) approaches 0.

Answer 13

An exponential grows facster than a polynomial (x-2).

Answer 14

faster*

Answer 15

so that entire expression approaches 0.

Answer 16

So that's because the exponent gets negative, and because of the property:
\[a ^{-n}=\frac{ 1 }{a ^{n} }\]
and because the e grows faster to infinity than the polynomial, it's 0?

Answer 17

yes. Look at e^x...as x------> negative infinity, e^x approaches 0...an it will outrun any polynomial as x approahes negative infinity.

Answer 18

e^x and e^(3x-1) is the same function when you want to look at end-behavior as x approaches negative infinity.

Answer 19

nice... Now I would need to study the growth of that function, meaning that I have to derivate it and study the behavior of the slopes.

Answer 20

I'm not clear what you need. You want to compute the derivative (via product rule)? You want any rel max/min?

Answer 21

study the sign of the derivative.

Answer 22

in order for obtaining rel max/min?

Answer 23

Yes.

Answer 24

Sure.

The derivative is f '(x) = 3(x-2)(e^(3x-1)) + e^(3x-1)

which can be factored as: e^(3x-1) [ 3(x-2) + 1}

so f ' (x) = 0 when 3(x-2) + 1 = 0, or x = 5/3.

Answer 25

\[f':f'(x)= e ^{3x+1}+(x-2)(e ^{3x+1})(3)\]

Answer 26

So, I'll wait till you check that out so that we're on the same page.

Answer 27

So the sign of f'(x) would be positive to the right and negative to the left.
meaning that it's a minimum.

Answer 28

I didnt do that yet..ket me see that for a moment.

Answer 29

Yes, x = 5/3 will be a rel min.

Answer 30

So f(5/3) = -134.4

Answer 31

I dont have a cal handy..so I'll take your word for it.

Answer 32

But (2,0) will be the x-intercept.

Answer 33

And (0, -2/e) is the y-intercept.

Answer 34

Which all makes sense, and as x----> + infinity, the graph goes to + infinity, as x = 5/3 is a rel min.

Answer 35

how did you get the intersections?

Answer 36

To find y - intercept(s), let x = 0.

To find x - intercept(s), let y = 0

Answer 37

work it out.

Answer 38

at y-intercept I got -2e^1

Answer 39

If x = 0, you get y = (0-2)e^(-1) = -2e^(-1) = -2/e

If y = 0, then x = 2 is the only solution (as e^(3x-1) can never equal to 0

Answer 40

You should have -2e^((-1)

Answer 41

agree?

Answer 42

yeah, I agree.

Answer 43

You are plugging in x = 0 into the actual function f(x).

Same for y = 0.

Answer 44

So now you have all the info to sketch the graph.

Answer 45

hmm, I'll need to see the concavity of that, so I'll need to do f''(x)

Answer 46

lol

Answer 47

You can do that if you want.

Answer 48

BTW, @Owlcoffee, f(5/3) is not -134.4
f(x) = (x - 2)e^(3x - 1)
f(5/3) = (5/3 - 2) * e^(3*5/3 - 1) = (-1/3) * e^4 = -18.2