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I did this the same way as the book example, and the class notes example, and the other problems in this section, which I got correct, but my answer for this one doesn't agree with the back of the book.
f has a limiting value 6 and passes through (0,3) and (1,4). Find the logistic function f.
[tex]f(x)=\frac{N}{1+Ab^{-x}}[/tex]
Assume that for small values of x, the function is exponential.
[tex]y=Ab^x[/tex]
[tex]4=Ab^1[/tex]
[tex]3=Ab^0[/tex] divide these equations:
[tex]\frac{4}{3}=b^1[/tex]
[tex]b=\frac{4}{3}[/tex]
Solve for A
[tex]f(x)=\frac{N}{1+Ab^{-x}}[/tex]
[tex]f(0)=3[/tex]
[tex]3=\frac{6}{1+Ab^{-0}}[/tex]
[tex]3=\frac{6}{1+A}[/tex]
[tex]3+3A=6[/tex]
[tex]3A=6-3[/tex]
[tex]A=1[/tex]
Answer:
[tex]f(x)=\frac{6}{1+(4/3)^{-x}}[/tex]
Back of book answer:
[tex]f(x)=\frac{6}{1+2^{-x}}[/tex]
f has a limiting value 6 and passes through (0,3) and (1,4). Find the logistic function f.
[tex]f(x)=\frac{N}{1+Ab^{-x}}[/tex]
Assume that for small values of x, the function is exponential.
[tex]y=Ab^x[/tex]
[tex]4=Ab^1[/tex]
[tex]3=Ab^0[/tex] divide these equations:
[tex]\frac{4}{3}=b^1[/tex]
[tex]b=\frac{4}{3}[/tex]
Solve for A
[tex]f(x)=\frac{N}{1+Ab^{-x}}[/tex]
[tex]f(0)=3[/tex]
[tex]3=\frac{6}{1+Ab^{-0}}[/tex]
[tex]3=\frac{6}{1+A}[/tex]
[tex]3+3A=6[/tex]
[tex]3A=6-3[/tex]
[tex]A=1[/tex]
Answer:
[tex]f(x)=\frac{6}{1+(4/3)^{-x}}[/tex]
Back of book answer:
[tex]f(x)=\frac{6}{1+2^{-x}}[/tex]
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