Thread: Linear Approximations

Hello, Everyone!

Any help would be greatly appreciated

Use linear approximation, i.e. the tangent line, to approximate as follows:
Let . The equation of the tangent line to at can be written in the form where is: ________
and where is:______
Using this, we find our approximation for is _________
NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer.


Im lost

Originally Posted by xcelxp

Hello, Everyone!

Any help would be greatly appreciated

Use linear approximation, i.e. the tangent line, to approximate as follows:
Let . The equation of the tangent line to at can be written in the form where is: ________
and where is:______
Using this, we find our approximation for is _________
NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer.


Im lost

for linear approximation, we use the formula:

f(x) ~= f(a) + f ' (a)(x - a)

where x is the value that you want to find f(x) for and a is a value close to x that you know the value of f(x) for.

note, that in your question they said use the line y = mx + b, this is the same thing, here b = f(a) and m = f ' (a) and (x - a) is our x

so let f(x) = sqrt(x)
we want to find sqrt(81.2), that is we want to find f(81.2)

do we know f(81.2)? no, but we know f(81) and 81 is very close to 81.2

so here our x = 81.2 and a = 81

so f(a) = f(81) = sqrt(81) = 9

f ' (x) = (1/2)x^(-1/2)
so f ' (a) = f ' (81) = (1/2)(81)^(-1/2) = (1/2)*(1/9) = 1/18

now we can just plug in the values into our equation:

using f(x) ~= f(a) + f ' (a)(x - a)

=> sqrt(81.2) = f(81.2) ~= f(81) + f ' (81)(81.2 - 81)
=> sqrt(81.2) = f(81.2) ~= 9 + (1/18)(0.2)
=> sqrt(81.2) = f(81.2) ~= 9 + 0.01111111111111111
=> sqrt(81.2) = 9.01111111

or for exact answer:

sqrt(81.2) ~= 9 + (1/18)(2/10) = 9 + 1/90 = 811/90

I am working out the problem, just one more question. Shouldnt b =9? This doesnt seem to be the right answer. Everything else worked out fine. Thanks!

Originally Posted by xcelxp

I am working out the problem, just one more question. Shouldnt b =9? This doesnt seem to be the right answer. Everything else worked out fine. Thanks!

let's find the equation of the tangent line at x = 81

f(x) = sqrt(x)
=> f ' (x) = (1/2)x^(-1/2)
=> the slope at 81 = f ' (81) = (1/2)(81)^(-1/2) = (1/2)(1/9) = 1/18
this is our m

now, when x = 81, y = sqrt(81) = 9

using the point slope form

y - y1 = m(x - x1)
=> y - 9 = (1/18)(x - 81)
=> y = (1/18)(x - 81) + 9
=> y = (1/18)x - 81/18 + 9
=> y = (1/18)x + 9/2

so 9/2 is our b