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April 20, 2014

April 20, 2014

Posted by **Lauren** on Wednesday, October 17, 2012 at 1:43pm.

Hint: The interval is [3,5]. What number is the center of that integral? And what is the function? There are many polynomials which answer this question correctly. PLease find one and explain why it is such a polynomial.

- Calculus II -
**Lauren**, Wednesday, October 17, 2012 at 3:27pmplease someone help this is due tomorrow and i don't know where to start with solving this!

- Calculus II -
**Jennifer**, Wednesday, October 17, 2012 at 3:34pm3^0.5 = 1.732

5^0.5 = 2.236

on [3,5],

3^x - 1.732 < 0.01

5^x - 2.236 < 0.01

3^x < 3^0.5 + 0.01

x < log(3) (1.742)

5^x < 5^0.5 + 0.01

x < log(5) (2.246)

For the first equation,

x < 0.50521360825

For the second equation,

x < 0.50275369436

x has to be lower than both of these numbers, and greater than 1/2, and it has to be a rational fraction.

So start trying numbers that are rational fractions slightly greater than 1/2

201/400 = 0.5025

so P(x) = x^(201/400) is one such polynomial

- Calculus II -
**Lauren**, Wednesday, October 17, 2012 at 3:38pmi dont understand why you raised the 3 and 5 to 1/2

- Calculus II -
**Steve**, Wednesday, October 17, 2012 at 4:26pmpolynomials have integer powers

f(3) = √3 = 1.732

f(5) = √5 = 2.236

Since this is calculus II, I assume you know about Taylor Polynomials. Let's use the polynomial for √x at x=4 (the midpoint of our interval)

√x = 2 + (x-4)/4 - ...

at x=3,5, we want |p(x)-f(x)| < .1

use enough terms to get that accuracy

if p(x) = 2,

p(3)-f(3) = 2-√3 = 0.26

p(5)-f(5) = 2-√5 = -.236

if p(x) = 2 + (x-4)/4 = 1-x/4,

p(3)-f(3) = 1.75 - √3 = 0.0179

p(5)-f(5) = 2.25 - √5 = 0.0139

Looks like a linear approximation fits the bill.

Just for grins, what happens if we use a parabola to approximate √x?

p(x) = 2 + (x-4)/4 - (x-4)^2/64

p(3)-f(3) = 0.0023

p(5)-f(5) = -0.0017

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