The main problem with every method discussed here is that if you make one wrong step, you get everything wrong. In all the above solutions,
a,
b and
c are very closely dependent on each other. Therefore a wrong answer in
a will result in a wrong answer for
b and
c. This is of course very bad for marks. So let's see how we can solve each variable one at a time.
=\frac{2x+1}{5x+3}\text{ and }g(x)=\frac{a}{x+b}+c)
By equating
)
with
)
we can obtain the solution, but again this will be prone to "one mistake leads to everything wrong" kind of situation as mentioned.
We can find
c by itself, using limits.
=\frac{2}{5}\mbox{, }\lim\limits_{x \to \infty }g(x)=c)
We can also find
b by itself, also using limits.
=\frac{1-2b}{3-5b}\mbox{, }\lim\limits_{x \to -b}g(x)=\infty)
=g(x)\mbox{ so }3-5b=0)
Finally we now we find
a. This should be less taxing since it uses no limits at all. Only derivatives.
=\frac{1}{(3+5x)^2}\mbox{, }g'(x)=-\frac{a}{(b+x)^2})
}=(3+5x)^2\mbox{, }\frac{1}{g'(x)}=-\frac{(b+x)^2}{a})
Keeping in mind that both sides are equal, we differentiate them two more times
}=50\mbox{, }\frac{d^2}{dx^2}\frac{1}{g'(x)}=-\frac{2}{a})
As you can see, the solutions for each variable are independent from each other. Now you do not have to worry about making a mistake from the first step. Even if one of them is wrong, the other two can still be right!
In conclusion:
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