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anonymous
 3 years ago
Find the linear approximation of the function
f(x) = sqrt of (1 − x) at a = 0. Use L(x) to approximate the numbers sqrt of 0.9
and sqrt of 0.99.
.
anonymous
 3 years ago
Find the linear approximation of the function f(x) = sqrt of (1 − x) at a = 0. Use L(x) to approximate the numbers sqrt of 0.9 and sqrt of 0.99. .

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I already found L(X) = 1/2X + 1. Idk what do do after that.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3\[\sqrt{1x} \approx \frac{1}{2}x+1\] \[\sqrt{0.9}=\sqrt{10.1} \approx \frac{1}{2}(0.1)+1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Srry but where did u get sqrt of 10.1 from if x = 0.9

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3I was trying to find out what x was

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3x is 0.1 since we want to approximate sqrt(0.9)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok so for sqrt of 0.99 it wud be \[\sqrt{0.99} = \sqrt{1 0.01} ~ 1/2 (0.01) + 1\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3You forgot the approximation symbol btw sqrt(10.01) and 1/2*(.01)+1 Also you should simplify 1/2*.01+1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the approximation is 0.995 right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So what is the general formula to write approxiamtions?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3\[\frac{1}{2} \cdot \frac{1}{100}+1 =\frac{1}{2} \cdot \frac{1}{100}+\frac{200}{200}=\frac{2001}{200}=\frac{199}{200} \text{ or } =.995\] So yes for sqrt(.99), that is a good approximation. To write linear approximations: Say we have a curve y=f(x) and we want to find the linear approximation at x=a for the numbers m. So we have: \[\text{ lines have this form: } y=mx+b\] Well we can find the general slope (aka derivative) for this curve. \[y'=f'(x)\] We actually wanted to know the slope at x=a That slope would in fact by y' evaluated at x=a So the slope of this tangent line is: \[y=f'(a)x+b\] We still need to find the yintercept, b. We do know a point on this line (a, f(a)) We can input this point in to find b. So we have: \[f(a)=f'(a) \cdot a +b\] \[f(a)f'(a) \cdot a =b \] So the tangent line to y=f(x) at x=a Or the equation we will use for linear approximations is: \[y=f'(a)x+f(a)f'(a) \cdot a \] or you might prefer to write it as: \[y=f'(a) \cdot (xa) +f(a)\] So anyways this is the linear approximation to y=f(x) for values near x=a. The approximations will get nastier the further your x is away from a. So we say: \[f(x) \approx f'(a)(xa)+f(a) \text{ for values of x near x }\] This is the general form for linear approximations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are u typing a reply? or is this a glitch?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3That slope m and that number m we want to use linear approximations on aren't the same. lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0WOW THANK U SOO MUCH FOR SUCH A thorugh explanation :D

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3Let's say we want to approximate n instead of m. So we have that \[n=f( * ) \approx f'(a)(*a)+f(a)\] What we input to approximate n depends on what the function is.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the star * symbol/ what does that represent in tha equation above

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3* is the number I chose to show you that we are not inputting n but we are inputting in a value so that the function is the same as n for some value of x. Here I just chose that value of x to be *.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh I SEE. This makes alot more sense now. Thank u soo much for putting so much time into heloing me with this question . I appreciate it.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3Np. I like linear approximations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank u :) I might need help with more just b/c sum I find hard to differentiate

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3k. Post em separately just in case I'm away. I have chores to do :(.
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