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The Curve y=e^x + 4e^(-2x) has one stationary point

(i) Find the x-coordinate of this point
(ii) Determine whether the stationary point is a maximum or a minimum point.
And
Is In 8e^(-2x) = 1 ?
Original post by PiersChoo
(i) Find the x-coordinate of this point
(ii) Determine whether the stationary point is a maximum or a minimum point.
And
Is In 8e^(-2x) = 1 ?


Where did you get that from?
Reply 2
I assume that the statement at the end is the result of your working, in which case it is wrong or it is the answer given in which case it is confusing and wrong.

I suspect a sign error in differentiating.


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Reply 3
Original post by nerak99
I assume that the statement at the end is the result of your working, in which case it is wrong or it is the answer given in which case it is confusing and wrong.

I suspect a sign error in differentiating.


Posted from TSR Mobile

the statement at the end is exactly my answer . Is it wrong?
Reply 4
Original post by nerak99
I assume that the statement at the end is the result of your working, in which case it is wrong or it is the answer given in which case it is confusing and wrong.

I suspect a sign error in differentiating.


Posted from TSR Mobile
'
i get that from pass year paper . Btw , the last statement is not part of the question . I am curious about that so i asked it .
Original post by PiersChoo
the statement at the end is exactly my answer . Is it wrong?


Yes it's wrong

That was why I asked where it came from
Reply 6
The differential of e2xe^{-2x} is 2e2x-2e^{-2x}. Setting up the equation equalling zero leads to a simple value for e^x. This all works a lot better if you post your working so that people can let you know where you are going wrong.

Your question was unclear in that you did not describe in any detail where you had issues.


Posted from TSR Mobile
(edited 9 years ago)
Reply 7
the derivative of eaxe^{ax} is aeaxae^{ax}

differentiate, make the eaxe^{-ax} term equal to 4aeax\displaystyle \frac{-4a}{e^{ax}}

Put the whole thing over a common denominator.

The numerator will factorise as a difference of 2 cubes.

Equate the denominator to zero and solve...

(the quadratic has no real roots, but i suggest you verify this by the discriminant)

For t he nature of the point, derive again, and shove in the value you solved for.

2nd derivative +ve => minimum

2nd derivative -ve => maximum
(edited 9 years ago)
Reply 8
Original post by Hasufel
the derivative of eaxe^{ax} is aeaxae^{ax}

differentiate, make the eaxe^{-ax} term equal to 4aeax\displaystyle \frac{-4a}{e^{ax}}

Put the whole thing over a common denominator.

The numerator will factorise as a difference of 2 cubes.

Equate the denominator to zero and solve...

(the quadratic has no real roots, but i suggest you verify this by the discriminant)

For t he nature of the point, derive again, and shove in the value you solved for.

2nd derivative +ve => minimum

2nd derivative -ve => maximum

Thanks
Reply 9
Original post by nerak99
The differential of e2xe^{-2x} is 2e2x-2e^{-2x}. Setting up the equation equalling zero leads to a simple value for e^x. This all works a lot better if you post your working so that people can let you know where you are going wrong.

Your question was unclear in that you did not describe in any detail where you had issues.


Posted from TSR Mobile

I got it now anyway thank you :>>

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