To find the first three non-zero terms in the Maclaurin series for the function f(x) = \frac{x^2}{x+1} , we start by writing the general form of the Maclaurin series:

f(x) = f(0) + x f^{\prime}(0) + \frac{x^2}{2!} f^{\prime\prime}(0) + \frac{x^3}{3!} f^{\prime\prime\prime}(0) + \cdots

We will evaluate each term by finding the function and its derivatives at x = 0 .

Evaluate f(0)

The first term in the series is f(0) .

f(0) = \frac{0^2}{0 + 1} = 0

Evaluate f^{\prime}(0)

The second term in the series is x f^{\prime}(0) . We have the first derivative from the previous part:

f^{\prime}(x) = \frac{x^2 + 2x}{(x + 1)^2}

Substituting x = 0 :

f^{\prime}(0) = \frac{0^2 + 2 \cdot 0}{(0 + 1)^2} = 0

Evaluate f^{\prime\prime}(0)

The third term in the series is \frac{x^2}{2!} f^{\prime\prime}(0). We have the second derivative from the previous part:

f^{\prime\prime}(x) = \frac{2}{(x + 1)^3}

Substituting x = 0 :

f^{\prime\prime}(0) = \frac{2}{(0 + 1)^3} = 2

Thus, the term is:

\frac{x^2}{2!} f^{\prime\prime}(0) = \frac{x^2}{2} \cdot 2 = x^2

Evaluate f^{\prime\prime\prime}(0)

The fourth term in the series is \frac{x^3}{3!} f^{\prime\prime\prime}(0). We need to find the third derivative f^{\prime\prime\prime}(x) by differentiating f^{\prime\prime}(x) :

f^{\prime\prime}(x) = \frac{2}{(x + 1)^3}

Using the chain rule:

f^{\prime\prime\prime}(x) = \frac{\differentialD}{\differentialD x} \left(2(x + 1)^{-3}\right) = 2 \cdot (-3)(x + 1)^{-4} = -6(x + 1)^{-4}

Substituting x = 0 :

f^{\prime\prime\prime}(0) = -6(0 + 1)^{-4} = -6

Thus, the term is:

\frac{x^3}{3!} f^{\prime\prime\prime}(0) = \frac{x^3}{6} \cdot (-6) = -x^3

Evaluate f^{(4)}(0)

The fifth term in the series is \frac{x^4}{4!} f^{(4)}(0). We need to find the fourth derivative f^{(4)}(x) by differentiating f^{\prime\prime\prime}(x) :

f^{\prime\prime\prime}(x) = -6(x + 1)^{-4}

Differentiating again:

f^{(4)}(x) = \frac{\differentialD}{\differentialD x} \left(-6(x + 1)^{-4}\right) = -6 \cdot (-4)(x + 1)^{-5} = 24(x + 1)^{-5}

Substituting x = 0 :

f^{(4)}(0) = 24(0 + 1)^{-5} = 24

Thus, the term is:

\frac{x^4}{4!} f^{(4)}(0) = \frac{x^4}{24} \cdot 24 = x^4

Conclusion

The first three non-zero terms in the Maclaurin series for f(x) are:

f(x) \approx x^2 - x^3 + x^4

Problem Bank - Maclaurin

Problem Bank - Maclaurin