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!
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.
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
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}} *)
Thank you for the input.
Thank you for the help!