What is the approximate value of f(0.98) using Euler's method?

daivinhtran
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Homework Statement



consider the function f(x) = aln(x+2). Given that f'(1) = a/3, what is the approximate value of f(0.98)?

Homework Equations


f(x1) = f(x0) + f'(x0)x(x1-x0)

The Attempt at a Solution


I solved it and get

f(.98) = aln(1+2) + (.098-1) = aln(3) - (.02)(a/3) <= not an answer

the bad thing is my teacher don't give me the increment interval. ..So I assumed it is .02
 
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daivinhtran said:

Homework Statement



consider the function f(x) = aln(x+2). Given that f'(1) = a/3, what is the approximate value of f(0.98)?





Homework Equations


f(x1) = f(x0) + f'(x0)x(x1-x0)


The Attempt at a Solution


I solved it and get

f(.98) = aln(1+2) + (.098-1) = aln(3) - (.02)(a/3) <= not an answer

the bad thing is my teacher don't give me the increment interval. ..So I assumed it is .02

That seems all right. Though you probably mean (0.98-1)(a/3) in you solution. Does 'not an answer' mean this is multiple choice? If so what are the possible answers?
 
Dick said:
That seems all right. Though you probably mean (0.98-1)(a/3) in you solution. Does 'not an answer' mean this is multiple choice? If so what are the possible answers?

A. (0.98-1)(a/3) + aln(2.98)
B. (a/3)aln(2.98) + .98
C. -.02a/3
D. (.98)ln(a/3)
E. (a/3) (.98) + aln(2.98)

none of them has aln3 like I solved
 
daivinhtran said:
A. (0.98-1)(a/3) + aln(2.98)
B. (a/3)aln(2.98) + .98
C. -.02a/3
D. (.98)ln(a/3)
E. (a/3) (.98) + aln(2.98)

none of them has aln3 like I solved

I don't care much for any of those choices either.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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