Is this the right answer? "Approximate: sqrt.(99.8)" Any and all help is greatly appreciated!
you know sqrt(100)
Yes! :)
so I suggest doing linerzation for f(x)=sqrt(x) at x=100
9.98
first step to do this is find the tangent line at x=100 \[y-f(100)=f'(100)(x-100) \\ y=f'(100)(x-100)+f(100)\]
this equation could be a pretty close approximation for values near x=100 and 99.8 is pretty close to 100
Right, that's what I did, I just wanted to make sure I did it right. It's been a bit lols;) I got 9.99 with the Linearization formula dealio. Is that close enough to the actual answer?
\[\sqrt{x} \approx f'(100)(x-100)+f(100) \text{ for values near } x=100\]
yep it should be in that ball park it should be something pretty close to 10 which is what you got
but I'm going to see what I get using the equation I have above
yep I got 999/100 as well which is 9.99
Sweet!! Thanks a zillion;)
putting it into the calculator gives me 9.989994... which is pretty damn close to what we got
Heck yes it is... sounds good to me:)
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