Linear Approximations

Hello, Everyone!

Any help would be greatly appreciated

Use linear approximation, i.e. the tangent line, to approximate http://hosted2.webwork.rochester.edu...372cabbfa1.png as follows:
Let http://hosted2.webwork.rochester.edu...54ba55ca11.png. The equation of the tangent line to http://hosted2.webwork.rochester.edu...910e13a9c1.png at http://hosted2.webwork.rochester.edu...66d300eca1.png can be written in the form http://hosted2.webwork.rochester.edu...9a0aa8b841.png where http://hosted2.webwork.rochester.edu...ddbff75931.png is: ________
and where http://hosted2.webwork.rochester.edu...dd497c3d61.png is:______
Using this, we find our approximation for http://hosted2.webwork.rochester.edu...372cabbfa1.png is _________
NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer.


Im lost:confused:

Quote:

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Originally Posted by xcelxp

Hello, Everyone!

Any help would be greatly appreciated

Use linear approximation, i.e. the tangent line, to approximate http://hosted2.webwork.rochester.edu...372cabbfa1.png as follows:
Let http://hosted2.webwork.rochester.edu...54ba55ca11.png. The equation of the tangent line to http://hosted2.webwork.rochester.edu...910e13a9c1.png at http://hosted2.webwork.rochester.edu...66d300eca1.png can be written in the form http://hosted2.webwork.rochester.edu...9a0aa8b841.png where http://hosted2.webwork.rochester.edu...ddbff75931.png is: ________
and where http://hosted2.webwork.rochester.edu...dd497c3d61.png is:______
Using this, we find our approximation for http://hosted2.webwork.rochester.edu...372cabbfa1.png is _________
NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer.


Im lost:confused:

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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!

Quote:

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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!

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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