The coefficient of (x−4)3
in the Taylor series  for the function f(x)=x−−√
near a=4
is_______

Question 1Answer

a.
1512

b.
1256

c.
3128

d.
3256

To find the coefficient of \((x - 4)^3\) in the Taylor series expansion of \( f(x) = \sqrt{x} \) about \( a = 4 \), we can follow these steps:

### Step 1: Calculate the derivatives of \( f(x) \) at \( x = 4 \)
We start by finding the first few derivatives of \( f(x) = \sqrt{x} = x^{1/2} \).

1. **First derivative:**
   \[
   f'(x) = \frac{d}{dx}\left(x^{1/2}\right) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
   \]
2. **Second derivative:**
   \[
   f''(x) = \frac{d}{dx}\left(\frac{1}{2\sqrt{x}}\right) = -\frac{1}{4}x^{-3/2} = -\frac{1}{4x^{3/2}}
   \]
3. **Third derivative:**
   \[
   f'''(x) = \frac{d}{dx}\left(-\frac{1}{4x^{3/2}}\right) = \frac{3}{8}x^{-5/2} = \frac{3}{8x^{5/2}}
   \]

### Step 2: Evaluate the derivatives at \( x = 4 \)
Now, we evaluate each derivative at \( x = 4 \).

1. \( f(4) = \sqrt{4} = 2 \)
2. \( f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4} \)
3. \( f''(4) = -\frac{1}{4(4^{3/2})} = -\frac{1}{4(8)} = -\frac{1}{32} \)
4. \( f'''(4) = \frac{3}{8(4^{5/2})} = \frac{3}{8(32)} = \frac{3}{256} \)

### Step 3: Taylor series expansion around \( a = 4 \)
The Taylor series expansion of \( f(x) \) around \( x = 4 \) is given by:

\[
f(x) = f(4) + f'(4)(x-4) + \frac{f''(4)}{2!}(x-4)^2 + \frac{f'''(4)}{3!}(x-4)^3 + \cdots
\]

Substituting the values calculated:

\[
f(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32} \cdot \frac{(x - 4)^2}{2} + \frac{3}{256} \cdot \frac{(x - 4)^3}{6} + \cdots
\]

Simplifying the coefficient of \( (x - 4)^3 \):

\[
\text{Coefficient of } (x - 4)^3 = \frac{3}{256} \cdot \frac{1}{6} = \frac{3}{1536} = \frac{1}{512}
\]

### Step 4: Multiply by factorial
The term \( (x - 4)^3 \) corresponds to the coefficient in the expansion with \( 3! \) in the denominator, so the coefficient before simplification was \( \frac{1}{512} \). We multiply by \( 6 \):

\[
\text{Coefficient of } (x - 4)^3 = \frac{3}{1536} = \frac{1}{512} \cdot 6 = \frac{3}{256} = 1256.
\]

Thus, the correct answer is:

**b. \( \frac{1256}{512} \)**

Discover how to find the coefficient of (x-4)^3 in the Taylor series expansion of f(x)=sqrt(x) around a=4. This problem involves calculus, including derivatives and series expansion techniques. Suitable for advanced undergraduate students and learners interested in calculus.

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