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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!

POSTED BY: Janet Gordon
4 Replies

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+b

In Mathematica this could be done in any of several ways. Here is one such.

Solve[{a*(ArcTan[0] + 2) == 2*E^b + 1,  2*E^(3*b) + 1 == Log[1] + 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[14]= {{a -> 3/2, b -> (2 Log[2])/3}} *)
POSTED BY: Daniel Lichtblau

Thank you for the help!

POSTED BY: Janet Gordon

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[3], b][[1]] 
sola = a /. Solve[ Limit[ f[a, x], x -> 0] == g[b, 0], a][[1]]

Unfortunately the values I get namely a=Log[4]/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 BY: Isaac Abraham

Thank you for the input.

POSTED BY: Janet Gordon
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