Math Problem Statement

In this question, we will calculate the Taylor Polynomial for f(x) = sqrt(x + 3) about x = 1.

Solution

Let's walk through the problem step by step to identify the mistakes and correct them. The task is to find the Taylor polynomial for f(x)=x+3f(x) = \sqrt{x + 3} about x=1x = 1.

Step 1: Calculating f(1)f(1)

The function is f(x)=x+3f(x) = \sqrt{x + 3}. Plugging in x=1x = 1:

f(1)=1+3=4=2f(1) = \sqrt{1 + 3} = \sqrt{4} = 2

So, f(1)=2f(1) = 2, which is correctly marked.

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

Using the chain rule, we differentiate:

f(x)=12(x+3)121=12(x+3)12f'(x) = \frac{1}{2} (x + 3)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2} (x + 3)^{-\frac{1}{2}}

Now, plug in x=1x = 1:

f(1)=12(1+3)12=1212=14f'(1) = \frac{1}{2} (1 + 3)^{-\frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

This is correct, and marked correctly.

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

Differentiate f(x)f'(x) again:

f(x)=ddx(12(x+3)12)f''(x) = \frac{d}{dx} \left( \frac{1}{2} (x + 3)^{-\frac{1}{2}} \right) f(x)=12(12)(x+3)32=14(x+3)32f''(x) = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) (x + 3)^{-\frac{3}{2}} = -\frac{1}{4} (x + 3)^{-\frac{3}{2}}

Now, plug in x=1x = 1:

f(1)=14(1+3)32=1418=132f''(1) = -\frac{1}{4} (1 + 3)^{-\frac{3}{2}} = -\frac{1}{4} \cdot \frac{1}{8} = -\frac{1}{32}

This is correct, and marked correctly.

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

Differentiate f(x)f''(x) again:

f(x)=ddx(14(x+3)32)f'''(x) = \frac{d}{dx} \left( -\frac{1}{4} (x + 3)^{-\frac{3}{2}} \right) f(x)=14(32)(x+3)52=38(x+3)52f'''(x) = -\frac{1}{4} \cdot \left( -\frac{3}{2} \right) (x + 3)^{-\frac{5}{2}} = \frac{3}{8} (x + 3)^{-\frac{5}{2}}

Now, plug in x=1x = 1:

f(1)=38(1+3)52=38132=3256f'''(1) = \frac{3}{8} (1 + 3)^{-\frac{5}{2}} = \frac{3}{8} \cdot \frac{1}{32} = \frac{3}{256}

This is also correct, and marked correctly.

Step 5: Constructing the Taylor Polynomial

The formula for the Taylor polynomial of degree 3 is:

T3(x)=f(1)+f(1)(x1)+f(1)2!(x1)2+f(1)3!(x1)3T_3(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3

Now, substitute the values:

T3(x)=2+14(x1)+13212(x1)2+325616(x1)3T_3(x) = 2 + \frac{1}{4}(x - 1) + \frac{-1}{32} \cdot \frac{1}{2}(x - 1)^2 + \frac{3}{256} \cdot \frac{1}{6}(x - 1)^3 T3(x)=2+14(x1)164(x1)2+1512(x1)3T_3(x) = 2 + \frac{1}{4}(x - 1) - \frac{1}{64}(x - 1)^2 + \frac{1}{512}(x - 1)^3

Errors in the Expression

The last two terms in the Taylor polynomial are incorrect:

  1. The coefficient for the (x1)2(x - 1)^2 term should be 164-\frac{1}{64}, but it's given as 1512\frac{1}{512}.
  2. The (x1)3(x - 1)^3 term coefficient is correct.

Correct Taylor Polynomial:

T3(x)=2+14(x1)164(x1)2+1512(x1)3T_3(x) = 2 + \frac{1}{4}(x - 1) - \frac{1}{64}(x - 1)^2 + \frac{1}{512}(x - 1)^3

Let me know if you need more details on any specific part!


Here are 5 questions that expand on this:

  1. How do you compute higher-order derivatives of f(x)=x+3f(x) = \sqrt{x+3}?
  2. Why do we divide by n!n! when constructing Taylor polynomials?
  3. How does the radius of convergence affect the accuracy of a Taylor polynomial?
  4. How would the polynomial change if we expanded around a different point, say x=0x = 0?
  5. Can you verify that the approximation for x=1.5x = 1.5 using this Taylor polynomial is accurate?

Tip: When expanding Taylor series, always check the factorial in the denominator, as it affects higher-order terms significantly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Taylor Series
Polynomial Approximation
Derivatives

Formulas

Taylor Polynomial of degree 3: T3(x) = f(a) + f'(a)(x - a) + f''(a)/2!(x - a)^2 + f'''(a)/3!(x - a)^3
f(x) = sqrt(x + 3)

Theorems

Taylor's Theorem

Suitable Grade Level

College Calculus