Math Problem Statement

Find the third order Taylor polynomial of f(x)= index 3 sqrt x at x=64.

Solution

We are tasked with finding the third-order Taylor polynomial of the function f(x)=x3f(x) = \sqrt[3]{x} around x=64x = 64.

The general form of the third-order Taylor polynomial for a function f(x)f(x) around x=ax = a is given by:

T3(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3T_3(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3

Here, a=64a = 64 and f(x)=x1/3f(x) = x^{1/3}. We'll need to compute f(a)f(a), f(a)f'(a), f(a)f''(a), and f(a)f'''(a).

Step 1: Compute f(a)f(a)

Since f(x)=x1/3f(x) = x^{1/3}:

f(64)=641/3=4f(64) = 64^{1/3} = 4

Step 2: Compute f(x)f'(x)

Using the power rule, the derivative of f(x)=x1/3f(x) = x^{1/3} is:

f(x)=13x2/3f'(x) = \frac{1}{3}x^{-2/3}

Now, evaluate f(x)f'(x) at x=64x = 64:

f(64)=13642/3=13116=148f'(64) = \frac{1}{3} \cdot 64^{-2/3} = \frac{1}{3} \cdot \frac{1}{16} = \frac{1}{48}

Step 3: Compute f(x)f''(x)

Now, find the second derivative of f(x)=13x2/3f(x) = \frac{1}{3}x^{-2/3}:

f(x)=ddx(13x2/3)=29x5/3f''(x) = \frac{d}{dx} \left( \frac{1}{3}x^{-2/3} \right) = -\frac{2}{9}x^{-5/3}

Evaluate f(x)f''(x) at x=64x = 64:

f(64)=29645/3=2911024=29216f''(64) = -\frac{2}{9} \cdot 64^{-5/3} = -\frac{2}{9} \cdot \frac{1}{1024} = -\frac{2}{9216}

Step 4: Compute f(x)f'''(x)

Finally, compute the third derivative of f(x)=29x5/3f(x) = -\frac{2}{9}x^{-5/3}:

f(x)=1027x8/3f'''(x) = \frac{10}{27}x^{-8/3}

Evaluate f(x)f'''(x) at x=64x = 64:

f(64)=1027648/3=102714096=10110592f'''(64) = \frac{10}{27} \cdot 64^{-8/3} = \frac{10}{27} \cdot \frac{1}{4096} = \frac{10}{110592}

Step 5: Construct the Third-Order Taylor Polynomial

Now, we substitute the values of f(64)f(64), f(64)f'(64), f(64)f''(64), and f(64)f'''(64) into the Taylor polynomial formula:

T3(x)=4+148(x64)+29216(x64)22!+10110592(x64)33!T_3(x) = 4 + \frac{1}{48}(x - 64) + \frac{-2}{9216} \frac{(x - 64)^2}{2!} + \frac{10}{110592} \frac{(x - 64)^3}{3!}

Simplifying:

T3(x)=4+148(x64)19216(x64)2+5110592(x64)3T_3(x) = 4 + \frac{1}{48}(x - 64) - \frac{1}{9216}(x - 64)^2 + \frac{5}{110592}(x - 64)^3

This is the third-order Taylor polynomial for f(x)=x3f(x) = \sqrt[3]{x} at x=64x = 64.

Would you like further details or any clarification?

Here are 5 questions for further exploration:

  1. How does a Taylor polynomial approximate a function?
  2. What are the conditions for a Taylor series to converge to the function it represents?
  3. Can higher-order terms in the Taylor series improve the accuracy of approximation?
  4. How would you find the fourth-order Taylor polynomial for the same function?
  5. What are the limitations of using a Taylor polynomial for approximation near singular points?

Tip: The accuracy of a Taylor polynomial depends on how far you are from the expansion point (here, x=64x = 64). Near the point, lower-order polynomials work well, but as you move further, higher-order terms become crucial for precision.

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Math Problem Analysis

Mathematical Concepts

Taylor Series
Derivatives
Polynomial Approximation
Power Functions

Formulas

Taylor polynomial: T_n(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3
Derivative of a power function: f'(x) = nx^(n-1)

Theorems

Taylor's Theorem

Suitable Grade Level

College Calculus or Advanced High School Calculus