How do you find the third-degree Taylor polynomial T3(x) for the function f(x) = tan^-1(x) centered at x = 7?

To find the third-degree Taylor polynomial T₃(x) for the function f(x) = tan^-1(x) centered at x = 7, we need to calculate the derivatives of f at that point and use them in the Taylor series expansion.

The Taylor polynomial of degree n for a function f centered at a point a is given by:

T_n(x) = f(a) + f'(a)(x – a) + f”(a)(x – a)²/2! + f”'(a)(x – a)³/3! + …

1. **Calculate the function value at a = 7:**

f(7) = tan^-1(7)

2. **Calculate the first derivative:**

f'(x) = 1 / (1 + x²)

Now plug in x = 7:

f'(7) = 1 / (1 + 7²) = 1 / 50

3. **Calculate the second derivative:**

f”(x) = -2x / (1 + x²)²

Again, plug in x = 7:

f”(7) = -2(7) / (1 + 7²)² = -14 / 2500 = -7 / 1250

4. **Calculate the third derivative:**

Using the quotient rule, we find:

f”'(x) = [2(1 + x²)²(1) – (-2x)(2)(1 + x²)(2x)] / (1 + x²)⁴

Calculating f”'(7) will give us the third derivative at x = 7.

Finally, substitute all these values into the Taylor polynomial formula:

T₃(x) = f(7) + f'(7)(x – 7) + f”(7)(x – 7)²/2! + f”'(7)(x – 7)³/3!

After calculating these derivatives and substituting, you will arrive at the third-degree Taylor polynomial T₃(x) for f(x) = tan^-1(x) centered at x = 7.