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bananna:
Hi can I have some help with this question please?
Thank you! :D

if the straight line y=mx is a tangent to the curve y=e^(x/2), find the exact value of m.

Hungry4Apples:

--- Quote from: bananna on May 15, 2017, 10:11:47 pm ---Hi can I have some help with this question please?
Thank you! :D

if the straight line y=mx is a tangent to the curve y=e^(x/2), find the exact value of m.

--- End quote ---

You would only be able to get a general solution unless you had a value of x. So i assume you just find the derivative function of e^(x/2) and use that as a sorta general gradient. Maybe m=0.5e^(x/2)

RuiAce:

--- Quote from: bananna on May 15, 2017, 10:11:47 pm ---Hi can I have some help with this question please?
Thank you! :D

if the straight line y=mx is a tangent to the curve y=e^(x/2), find the exact value of m.

--- End quote ---

RuiAce:

--- Quote from: MisterNeo on May 15, 2017, 11:05:28 pm ---I got
(That's an index btw)
(Image removed from quote.) Not sure if I did it correctly, but yeah.

--- End quote ---
There is a circular argument here and an assumption without proof which I don't quite understand.

You firstly substitute in \(m=2\) and then conclude \(m\) is also something else. Explain?

RuiAce:

--- Quote from: MisterNeo on May 15, 2017, 11:23:00 pm ---I was trying to prove that m=1/2e^1/2x by substituting m=2 to find x to sub into the equation again to show that they were equal. It doesn't really work mathematically I guess...  :-X

--- End quote ---
The aim is to show that m is a constant. Leaving x in there defeats the purpose, which is why for this particular question we needed two equations to solve simultaneously.

Substituting something in to prove that it's true is basically assuming what you're trying to prove. It's like proving x = 4 is a solution to \(x^2-4x=0\) by substituting it in, but then claiming it's the only solution. (Take out: Never assume what you're trying to prove)

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