Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=x e^{x}, \quad n=4 $$

Short answer

The Maclaurin polynomial of degree 4 for the function is \(f(x) = x + x^2 + \frac{x^3}{2} + \frac{x^4}{6}\).

Step-by-step solution

Understand the Taylor Series

The Taylor Series of a function is an infinite sum of terms based on the values of the function's derivatives at a single point. For the function \( f \) differentiable at the point 0, the Maclaurin series is: \[ f(x) = f(0) + f'(0)x + \frac{f''(0) x^2}{2!} + \frac{f'''(0) x^3}{3!} + \frac{f''''(0) x^4}{4!} + ... \] where \(f'(0), f''(0), f'''(0), f''''(0)\) denotes the first, second, third and fourth derivatives of \(f\) at 0.

Calculate the derivatives

Next, differentiating \(f(x) = x e^{x}\) repeatedly up to 4 times gives: \(f'(x) = e^{x} + x e^{x}, f''(x) = 2e^{x} + x e^{x}, f'''(x) = 3e^{x} + x e^{x}, f''''(x) = 4e^{x} + x e^{x}\). Then calculate the value of each derivative at 0. We get \(f(0) = 0, f'(0) = 1, f''(0) = 2, f'''(0) = 3, f''''(0) = 4.\)

Substitute the values into the Maclaurin series

Substitute these values into the Maclaurin series up to degree 4, we get: \(f(x) = 0 + 1x + \frac{2x^2}{2!} + \frac{3x^3}{3!} + \frac{4x^4}{4!}\)

Simplify the expression

Simplifying this expression, we have: \(f(x) = x + x^2 + \frac{x^3}{2} + \frac{x^4}{6}\).