Calculus ...help plz....!!!

We wish to approximate sqrt(20) , but 20 isn't very close to a perfect square . Find sqrt(80) using tangent line approximation and use your answer to approximate sqrt(20) .

y = √x
9 = √81

So that's your point, (81, 9).

y = x^(1/2)
y' = (1/2) * x^(-1/2)
y' = 1/(2√x)
y' = 1/18

That's your slope. Put it all together.

y = mx + b
b = y - mx
b = 9 - (1/18)(81)
b = 9 - 9/2
b = 9/2

So the tangent line is: y = x/18 + 9/2

So an approximate for √80 is...

√80 ≈ 80/18 + 9/2
√80 ≈ (80 + 81)/18
√80 ≈ 161/18 = 80.9444... (repeating 4)

Compared to the real value: http://www.wolframalpha.com/input/?i=sqr…

Note that √80 = √4√20 = 2√20. So √20 = √(80)/2. Then...

√20 ≈ 161/36 = 4.472222... (repeating 2)

Compared to the real value: http://www.wolframalpha.com/input/?i=sqr…

Heh, whoops...at least it was an obvious mistake...

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√80 = 2√20
√20 = √80/2

y = √x
y' = 1/(2√x) = m

y - y0 ≈ m(x-x0)
√80 - √81 ≈ m(80-81)
√80 ≈ 9 - m = 9 - 1/(2√81) = 9-1/18 = 161/18

√20 = (1/2)√80 ≈ (1/2)(161/18) ≈ 4.472 (correct to the 3rd decimal place)

f(x) = √x

f(80) = √80 = 4√5

f ' (x) = 1 /(2√x)

f ' (80) = 1/ (2√80) = 1/(8√5) = √5/40

L(x) = f(80) + f ' (80) (x - 80)

L(x) = 4√5 +(√5/40)(x - 80)

L(x) = 4√5 + √5 x/40 - 2√5

L(x) = 2√5 + √5 x/40

√20 = L(20) = 2√5 + √5(20)/40

= 2√5 + √5/2

= 1/2 (5√5)

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