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# Calculus: Continuity problem.

Posted 10 years ago
 Find the values of a and b that make the given function continuous. f(x)={a(tan^-1 x +2), if x<0, 2e^bx +1, if 0<(or equal to) x <(or equal to) 3, ln(x-2) +x^2, if x>3  I cannot get WolframAlpha to understand this question and everyone I have asked for help also has not been able to figure it out, does anyone here know how to solve this or know how to make Wolfram understand it? Hopefully I worded this appropriately to actually get help this time. Thanks!
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Posted 10 years ago
 Use input that forces compatibility at the two junctures. In other places these functions are continuous. solve a*(atan(0)+2)=3, 2*e^(3b)+1=9 for a, b http://www.current.wolframalpha.com/input/?i=solve+a*%28atan%280%29%2B2%29%3D3%2C+2*e^%283b%29%2B1%3D9+for+a%2C+bIn Mathematica this could be done in any of several ways. Here is one such. Solve[{a*(ArcTan + 2) == 2*E^b + 1, 2*E^(3*b) + 1 == Log + 3^2}, {a, b}, Reals] This is more elaborate but allows one more flexibility in handling. f1[a_, b_, x_] := a*(ArcTan[x] + 2) f2[a_, b_, x_] := 2*Exp[b*x] + 1 f3[a_, b_, x_] := Log[x - 2] + x^2 Solve[{f1[a, b, 0] == f2[a, b, 0], f2[a, b, 3] == f3[a, b, 3]}, {a, b}, Reals] (* Out= {{a -> 3/2, b -> (2 Log)/3}} *) 
Posted 10 years ago
 Thank you for the help!
Posted 10 years ago
 I'm attempting to answer for the "Mathematica" listed above. f[a_, x_] := a (ArcTan[x] + 2);(*x<0*) g[b_, x_] := 2 Exp[b x] + 1; (*0<=x<=3*) h[x_] := Log[x - 2] + x^2; (*x>3*) solb = b /. Solve[ Limit[g[b, x], x -> 3] == h, b][] sola = a /. Solve[ Limit[ f[a, x], x -> 0] == g[b, 0], a][] Unfortunately the values I get namely a=Log/3=0.46 and b=3/2 do not result in the three pieces having the same slope at the "joins". Enforcing the same slope at the two joins gave a=1.18 and b=0.59. Therefore I don't think there is a solution. Maybe someone else will post a better solution.
Posted 10 years ago
 Thank you for the input.