Calculus II

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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.

  • Calculus II -

    please someone help this is due tomorrow and i don't know where to start with solving this!

  • Calculus II -

    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

  • Calculus II -

    i don't understand why you raised the 3 and 5 to 1/2

  • Calculus II -

    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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