Posted by Lauren on Wednesday, October 17, 2012 at 1:43pm.
please someone help this is due tomorrow and i don't know where to start with solving this!
3^0.5 = 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
i dont understand why you raised the 3 and 5 to 1/2
polynomials 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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