anonymous
  • anonymous
Find a and b such that the derivative of f exists everywhere, if f(x)= {x^2+3, (x < or =1) and ax^3+bx+4, x>1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
this is much easier than it looks take the derivative of both replace \(x\) by \(1\) and set them equal
anonymous
  • anonymous
so 2x and a3x^2+b 2(1)=1 a3(1)^2+b=1 ?
anonymous
  • anonymous
lost.

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anonymous
  • anonymous
still trying to figure out if I am on the right track. I am lost right now. No idea what to do after I find their derivatives never had an a and b in the equation like this.
anonymous
  • anonymous
@Fakshon as far as derivatives are concerned...you got that part right.. 2x and 3a*x^2+b now you have to set: 3a*x^2+b=2x and put x=1
anonymous
  • anonymous
so you got one equation with a and b?
anonymous
  • anonymous
still lost?Just nod...we can go to the begining
anonymous
  • anonymous
you function is defined in 2 different ways depending on the interval. The limit point of the intervals is x=1. Independently, the derivative exist on each interval separatly, but for them to exist in all R it is needed that they are equal in this limiting point x=1. So like @satellite73 said, find first each derivative. Later set x=1 and put them equal. What is left is to choose a and b that make the equality happen. That's it
anonymous
  • anonymous
@myko we get one equation in a and b by differentiating both parts and setting them equal at x=1...the other equation is obtained from the fact that function is continuous at x=1...since continuous functions are differentiable and vice versa... so we can equalize x^2+3 and ax^3+bx+4 at x=1..so that the function is continuous.. now two equations..two unknowns.. can be worked out
anonymous
  • anonymous
right

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