Find the inverse. (1 Viewer)

[miithun]

Find the inverse of
y=xlnx

Thank you

 

[miithun]

I don't think you know what you are doing.
You came to the question in the end because you didn't swap the x's and y's.
Anyone else?

 

[hscishard]

I know I rofled.

 

[miithun]

Lmfao

 

[Aquawhite]

By the fundamental log property that:





Here's the graph that you will most likely need to draw in the second part which of course you can work out using calculus:

 

[tommykins]

e^ylny =/= y^2.

 

[Trebla]

By the fundamental log property that:





Here's the graph that you will most likely need to draw in the second part which of course you can work out using calculus:

I don't think that is correct. I'm not sure what you did, but the two functions are not symmetric about y = x.

Is this the full question or is it asking you to evaluate the inverse at a certain number (which can be done without explicitly finding the inverse)?

 

[Aquawhite]

Well I think it's right anyway

 

[tommykins]

Well I think it's right anyway

No.

y = xlnx = ln(x^x)

hence e^y = x^x, not x^2.

 

[hscishard]

» | Documentation | Properties | Definition


<HR style="TOP: auto" class=bot><HR class=top>Result:

» | Documentation | Properties | Definition


<HR style="TOP: auto" class=bot>
<HR class=top>Plot:

Lol. Doubt that's HSC stuff.

 

[Aquawhite]

No.

y = xlnx = ln(x^x)

hence e^y = x^x, not x^2.

Well, would you care to provide us with a full solution please?

 

[Aquawhite]

» | Documentation | Properties | Definition


<HR style="TOP: auto" class=bot><HR class=top>Result:

» | Documentation | Properties | Definition


<HR style="TOP: auto" class=bot>
<HR class=top>Plot:

Lol. Doubt that's HSC stuff.

I just looked at that and I was like "eh?, Dunno where there going with that..."

 

[tommykins]

Well, would you care to provide us with a full solution please?

I don't know how to do the initial question.

but alogb = log(b^a)

 

[Aquawhite]

I don't know how to do the initial question.

but alogb = log(b^a)

Yes, I recognise that... I was trying to adapt the idea by basically having the y represent just any number... but just testing a few on the calculator I see that it was wrong. At least I gave it a go.

 

[kaz1]

Doesn't look like HSC, tried it on Maple and got y/LambertW(y)

 

[miithun]

I will provide the full question.
a) Consider the function f(x) = xlnx
i) Show that y=f(x) has minumum turning point at [1/e, -1/e] [2marks]
ii) Hence sketch the curve y=f(x) for x>= 1/e
iii) Draw the sketch of the inverse function of y=f(x) for x>= 1/e

Sorry guys i just realised you don't need to find inverse function because you can just refelct the graph of part ii) over the line y=x

 

[tommykins]

Yes, I recognise that... I was trying to adapt the idea by basically having the y represent just any number... but just testing a few on the calculator I see that it was wrong. At least I gave it a go.

Yes I know, not a problem. Minor technicalities.

 

[kooltrainer]

the inverse is : y^y = e^x
dont need to make y the subject..coz its the same thing..