g C bF D SolutionTo find g x C bxD Given g027 g 112 For g

Solution

To find g (x)= C ×b^x+D

Given g(0)=27, g (1)=12

For g (0), substitute x=0 in g (x)

g (0)= C b⁰+D= 27

C+D=27

g (1)= C b¹+D=12

Cb+D=12 , since there are three constants and only 2 equation, we can\'t solve it by simultaneous equations method.

Another condition is g(x)-> 1 as x->infinity

For this b should be a fraction of a form 1/p where p is a constant, only then b^x=1/p^x , this is necessary because if x was in the numerator, g wouldn\'t tend to 1, for g to tend to 1 , when x would tend to infinity , D should be 1, because

When x->infinity, 1/p^x =1/p^infinity->1/infinity =0

Then C×0+D= 1 thus D=1

C+D=27

C=27-1= 26

26b+ D=12

26b+1=12

b= 11/26

b=1/p= 11/26

p=1/(11/26)

g (x)= 26 ×1/(11/26)^x+ 1