My Math Forum need help with linear approximation problem

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 November 1st, 2011, 03:42 PM #1 Newbie   Joined: Nov 2011 Posts: 1 Thanks: 0 need help with linear approximation problem Use linear approximation to estimate the value of f(3.1), given that f(3)=9 and f '(x) = (5x^2) / (sqrt(x^3-2)) Please help !
 November 2nd, 2011, 12:32 AM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: need help with linear approximation problem f(x+dx) = f(x) + ( f ' (x) )*dx approximately x+dx = 3.1 x =3 dx =0.1 f(x) = f(3) = 9 compute f ' (x) = f ' (3) thanks to the given formula then compute f(x) + ( f ' (x) )*dx
 November 2nd, 2011, 01:33 AM #3 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 Re: need help with linear approximation problem $\frac{dy}{dx}=\frac{5x^2}{\sqrt{x^3-2}} \\ \Rightarrow y=\int \frac{5x^2}{\sqrt{x^3-2}} dx$ Let $u=x^3-2$. Then: $\frac{du}{dx}=3x^2 \Leftrightarrow dx=\frac{du}{3x^2}$ Integral becomes $\int \frac{5x^2}{\sqrt{x^3-2}} dx= \frac{5}{3}\int \frac{1}{\sqrt{u}}du = \frac{10 \sqrt{u^}}{9}= \frac{10 \sqrt{x^3-2}}{9}$ $f(x)=y \\ \Rightarrow f(x)=\frac{10 \sqrt{x^3-2}}{9} \\ \Rightarrow f(3.1)\approx 9.1146774591146428751161836275840270305$
 November 2nd, 2011, 01:35 AM #4 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 Re: need help with linear approximation problem Sorry I went wrong with the last step.$f(3.1)\approx 3.0651364942519383$
 November 2nd, 2011, 03:11 AM #5 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: need help with linear approximation problem Hello Etyucan, you forgot the integration constant. The correct analytical result is : 9+(10/3)sqrt[(x^3)-2] - (10/3)sqrt[(3^3)-2] = (10/3)sqrt[(x^3)-2] -(23/3) = 9.90572329283767 for x=3.1 But isn't what pdeep needs to answer to his problem. Linear approximation is much simpler and suffisant : the resul is close to the exact value.

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