anonymous
  • anonymous
Find the smallest value of a so that the function y=e^(2x) *x^2 * a*e^(2x)+3 would be growing no matter what the x value is.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
The answer given is 1.
anonymous
  • anonymous
Sorry, I made a mistake. The function is y=e^(2x) *x^2+a*e^(2x)+3
experimentX
  • experimentX
what does "would be growing no matter what the x value is" mean??

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anonymous
  • anonymous
The function gets bigger not smaller.
experimentX
  • experimentX
a>0
anonymous
  • anonymous
Oh, alright. Thanks!
experimentX
  • experimentX
seems to be growing for a=0.001
experimentX
  • experimentX
wait thats not correct
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=+y%3De%5E%282x%29+%28x%5E2+-+25%29+%2B3 seems to me that the function is always growing after certain interval of time
experimentX
  • experimentX
I think a=1 is the right answer ...
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=+y%3De%5E%282x%29+%28x%5E2%2B1%29+%2B3
anonymous
  • anonymous
But how do you get it?
experimentX
  • experimentX
for a=0, we have y=e^(2x) (x^2) +3 and this (x^2) is less than 1 between 0 to 1 and is decreasing the function.
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=+y%3De%5E%282x%29+%28x%5E2%2B1%29+%2B3
anonymous
  • anonymous
Oh, alright.
experimentX
  • experimentX
@ErkoT you still there ??? still I am quite incorrect
experimentX
  • experimentX
y=e^(2x) (x^2) +3 increases between 0 to 1 it decreases on -1 to 0
experimentX
  • experimentX
ah ... i found the best way, take derivative of e^(2x) (x^2+a) and it must be > 0, and find the value of a as x->-0

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