ax^2 + 1/3, x> 1 f(x)= bx -10/3, x<1 If the function is differentiable, find the sum of a + b I got that a+ 1/3 = b - 10/3 Where do I go from there?
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you did the f(x) is continuous part ... need to do this, too, to get the second equation. $\displaystyle \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x)$
Ok, then I get 2a = b, so do I plug that back in to the original equations that were set equal to one another? EDIT: I ended up getting 11
Last edited by CalcGeek31; Jan 5th 2009 at 01:54 PM.
Originally Posted by CalcGeek31 Ok, then I get 2a = b, so do I plug that back in to the original equations that were set equal to one another? EDIT: I ended up getting 11 Mr F says: Correct. For the record: Now you solve the following equations simultaneously for a and b: $\displaystyle a + \frac{1}{3} = b - \frac{10}{3}$ .... (1) $\displaystyle 2a = b$ .... (2) Then you calculate the value of a + b. You get what you got.
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