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I have done this question but when I draw the graph of Lo(x) I dont get a linear function...I get some weird looking graph kinda U shaped. its above f(x) so I think I have the wrong answer.. the question is find the linear approximation of x/sqrt(X+1) at x=0, use the linearization to approximate f(0.03)

my answer, linear approx= x/(sqrt(x+1))+(2+x)/(2 (1+x)^(3/2))(x-0)

The Lo graph isnt linear though or atleast maybe my scale is wrong...I really need help. Am I wrong? what am I doing wrong?

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What question are you talking about? What is Lo(x) and f(x)? –  Chris Taylor Nov 27 '11 at 0:19
But you know how to find the tangent line of $\dfrac{x}{\sqrt{x+1}}$ at $x=0$, don't you? –  J. M. Nov 27 '11 at 0:24
Yes. I told you my answer that I got, the first linear approximation of the function x/sqrt(X+1) at x=0 is x/(sqrt(x+1))+(2+x)/(2 (1+x)^(3/2))(x-0) Lo is the very first linear approximation. –  Raynos Nov 27 '11 at 0:34
Yes, you did it wrong. Probably because you keep confusing the independent variable in the tangent line with the independent variable of your function. Here's the algorithm you should follow: you have an $f(x)$ and a given $x_0=0$; take the derivative $f^\prime(x)$, and evaluate it as $m=f^\prime(x_0)$. This $f^\prime(x_0)$ should be a number, not a function of $x$. The line you need is $y=m(x-x_0)+f(x_0)$. Again, $f(x_0)$ is a number, and not a function of $x$. –  J. M. Nov 27 '11 at 0:49
yeah but i do sub in for x as 0.03 when i do the calculations..does me putting it as a function form mess it up? holy crap im so confused now.....then what happens when i need to approximate 0.03 using that approximation...could you please show me the solution so i can get the grasp? –  Raynos Nov 27 '11 at 0:59

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