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Exact coordinates of M (The minimum point)

I have been doing past paper questions whole day, and each time i come across a problem i get the best help from this site, i only have two more questions and they are hard, in the first part of this question i have curve y = xe2x and it has a Minimum point M, how can i find M..... im thinking differentiation using product rule but i get stuck at

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Re: Exact coordinates of M (The minimum point)

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Originally Posted by

**johnreal** I have been doing past paper questions whole day, and each time i come across a problem i get the best help from this site, i only have two more questions and they are hard, in the first part of this question i have curve y = xe2x and it has a Minimum point M, how can i find M..... im thinking differentiation using product rule but i get stuck at

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Please keep in mind that $\displaystyle \displaystyle{\frac{de^x}{dx} = e^x}$

$\displaystyle f(x)=x \cdot e^{2x}~\implies~f'(x)=e^{2x}+x \cdot e^{2x} \cdot 2 = e^{2x}(1+2x)$

Now continue!

Re: Exact coordinates of M (The minimum point)

Thanks but how is de^x/dx=e^x

Re: Exact coordinates of M (The minimum point)

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Originally Posted by

**johnreal** Thanks but how is de^x/dx=e^x

Pardon?

In case you really don't know this have a look here (especially #3.): Exponential function - Wikipedia, the free encyclopedia

Re: Exact coordinates of M (The minimum point)

So, e is a specific #?, i think i understand but how do i find M, think im totally lost.

Re: Exact coordinates of M (The minimum point)

why are you attempting a problem involving an exponential function of base "e" w/o knowing anything about the base number, much less anything about the derivative of exponential functions in general?

Re: Exact coordinates of M (The minimum point)

Ask my teacher, i just wanna understand my homework, if someone could explain

Re: Exact coordinates of M (The minimum point)

I think you need to ask your teacher that question. I find it difficult to believe that your teacher would assign a problem in finding an extremum for a function with an exponential factor w/o the requisite instruction on the basics of finding the derivative of such a function.

Meanwhile, I recommend you research how to find derivatives of exponential functions.

Re: Exact coordinates of M (The minimum point)

Quote:

Originally Posted by

**johnreal** I have been doing past paper questions whole day, and each time i come across a problem i get the best help from this site, i only have two more questions and they are hard, in the first part of this question i have curve y = xe2x and it has a Minimum point M, how can i find M..... im thinking differentiation using product rule but i get stuck at

Attachment 23395

I may be perceiving this the wrong way, but if your function is x(e^2x), then all you need is to take the derivative, find where it is equal to zero, then either take the second derivative to determine concavity or do as my calc teacher taught us and make a "sign" chart which entails taking values of the first derivative to the right and left of the place where the first derivatives value is zero and finding if they are positive or negative, which then allows you to determine if it's a max or min.

So essentially take the derivative:

dy/dx= e^2x + 2xe^2x

(the first two in 2xe^2x comes from the chain rule)

(The derivative of e^(almost any variable) is equal to itself + the chain rule.)

Then set it equal to zero.

e^2x + 2x(e^2x) = 0

Then solve:

(e^2x)(1+2x) = 0

1+2x = 0

x=-.5

To check that it is a minimum plug some values in to the derivative to the left and right of -.5. To make it simple I'm going to use -1 and 0.

e^(2*-1) + 2(-1)(e^(2*-1) = -.135 Which is negative, meaning that the function is sloping down from the left.

e^(2*0) + 2(0)(e^(2*0) = 1 Which is positive meaning that the function is sloping up at this point.

Since we know the function is sloped downward on the left of x= -.5 and sloped upward to the right of x=-.5 we can determine that x=-.5 is a minimum.

Which agrees with what our calculator determined below.

Another way to do this is to graph it... and find the minimum.

Min at x=-.5000011

Min of y=-.1839397

Looking at some of khanacademy's videos may help also if you are having a hard time understanding.

P.S. Does anyone know why the derivative of e^(whatever variable) is almost always itself + the chain rule?

Does is have anything to do with e = lim n->infinity (1+1/n)^n?

Re: Exact coordinates of M (The minimum point)

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Originally Posted by

**Xeritas** P.S. Does anyone know why the derivative of e^(whatever variable) is almost always itself + the chain rule?

Does is have anything to do with e = lim n->infinity (1+1/n)^n?

let $\displaystyle y = e^u$ where $\displaystyle u$ is a function of $\displaystyle x$

$\displaystyle \ln{y} = u$

$\displaystyle \frac{d}{dx} \left(\ln{y} = u\right)$

$\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} = \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = y \cdot \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = e^u \cdot \frac{du}{dx}$

Re: Exact coordinates of M (The minimum point)

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Originally Posted by

**skeeter** let $\displaystyle y = e^u$ where $\displaystyle u$ is a function of $\displaystyle x$

$\displaystyle \ln{y} = u$

$\displaystyle \frac{d}{dx} \left(\ln{y} = u\right)$

$\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} = \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = y \cdot \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = e^u \cdot \frac{du}{dx}$

Thanks, although I was thinking more conceptually/visually rather than algebraically.

Re: Exact coordinates of M (The minimum point)

Thanks everyone, i looked up alot on exponential functions and i think i grasp it enough, also special thanks to Xeritas you confirmed my answer for me.