Calculus II
posted by Lauren .
Fed loves polynomials with rational coefficients and only such polynomials.Suppose f(x)=square root of x. Find a polynomial P(x) that Fred will adore so that, for any x in the interval [3,5], the difference between P(x) and f(x) is less than .01
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.

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 don't 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 + (x4)/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 + (x4)/4 = 1x/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 + (x4)/4  (x4)^2/64
p(3)f(3) = 0.0023
p(5)f(5) = 0.0017