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
In case you really don't know this have a look here (especially #3.): Exponential function - Wikipedia, the free encyclopedia
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.
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
(e^2x)(1+2x) = 0
1+2x = 0
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?