Can someone please show me where I went wrong in solving the following problem.

Find the local linear approximation of the function f(x)=sqrt(1+x) at x0=0, and use it to approximate sqrt(0.9) and sqrt(1.1)

Question

Can someone please show me where I went wrong in solving the following problem.

Find the local linear approximation of the function f(x)=sqrt(1+x) at x0=0, and use it to approximate sqrt(0.9) and sqrt(1.1)

anonymous · Answer

your derivative is $\frac{1}{2\sqrt{x+1}}$

babyslapmafro · Answer

Here is what I did...
Approximation:  sqrt(1)
x=0
f(x)=sqrt(1+x)=1
f'(x)=1/[2sqrt(1+x)]=1/2 
point (0,1)

anonymous · Answer

at $x=0$ your slope is $\frac{2(x^2+1)-2x(2x)1}{2}$ and your line is 
$$y=\frac{1}{2}x+1$$

anonymous · Answer

what on earth???

babyslapmafro · Answer

y-1=.5(x-0)

y=.5(.9-0)+1=1.45
y=.5(1.1-0)+1=1.55

anonymous · Answer

i meant your slope is $\frac{1}{2}$

anonymous · Answer

so you get 
$$y=\frac{1}{2}x+1$$ now you want to approximate $\sqrt{.9}$ so if $x+1=.9$ then $x=-.1$

anonymous · Answer

i see the mistake, you used $x=.9$ not $x=-.1$

babyslapmafro · Answer

Ok, so I was plugging in the wrong value for x

anonymous · Answer

yeah, it is $1+x=.9$ and $1+x=1.1$ and so $x=-.1$ and $x=.1$ respectively

babyslapmafro · Answer

Ok, then why did they bother giving me the value for x-base 0? It is unnecessary correct?

babyslapmafro · Answer

I think that value kinda confused me