Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\tan x, \quad n=3 $$

Short answer

The Maclaurin polynomial of degree 3 for the function \( f(x) = \tan x \) is \( x + \frac{x^3}{3} \).

Step-by-step solution

Compute the function value and derivative at \( x = 0 \)

Compute the function value at \( x = 0 \): \( f(0) = \tan 0 = 0 \). Calculate the first derivative of \( f(x) = \tan x \): \( f'(x) = \sec^2 x \), and then evaluate at \( x = 0 \): \( f'(0) = \sec^2 0 = 1 \).

Compute the second derivative

The second derivative of \( f(x) \) is given by the derivative of \( f'(x) \). Thus, \( f''(x) = 2\tan x \cdot \sec^2 x \) and when evaluated at \( x = 0 \), we get \( f''(0) = 0 \).

Compute the third derivative

The third derivative of \( f(x) \) is then \( f'''(x) = 2\sec^2 x + 4\tan^2 x \sec^2 x \). Evaluate this at \( x = 0 \) to get \( f'''(0) = 2 \).

Construct the Maclaurin Polynomial

Combine these results to form the Maclaurin polynomial of degree 3: \( P_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} = 0 + 1x + 0x^2 + \frac{2x^3}{6} = x + \frac{x^3}{3} \).