Let f(x) = ex + x2 for real x. From among the following, choose the Taylor series approximation of f(x) around x = 0, which includes all powers of x less than or equal to 3, A. 1 + x + x2 + x3 B. 1 + x + \[\frac{3}{2}\]x2 + x3 C. 1 + x + \[\frac{3}{2}\]x2 + \[\frac{7}{6}\]x3 D. 1 + x + 3×2 + 7×3

x2 + x3″ option3=”1 + x + \[\frac{3}{2}\]x2 + \[\frac{7}{6}\]x3″ option4=”1 + x + 3×2 + 7×3″ correct=”option3″]

The correct answer is $\boxed{1 + x + \frac{3}{2}x^2 + x^3}$.

The Taylor series of a function $f(x)$ centered at $x=0$ is given by

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$

where $f^{(n)}(0)$ is the $n$th derivative of $f$ evaluated at $x=0$.

In this case, $f(x) = ex + x^2$, so

$$f'(x) = e^x + 2x$$

$$f”(x) = e^x + 2$$

$$f”'(x) = e^x$$

and so on.

Evaluating these derivatives at $x=0$ gives

$$f^{(0)}(0) = 1$$

$$f^{(1)}(0) = 1 + 2 = 3$$

$$f^{(2)}(0) = 1 + 2 = 3$$

$$f^{(3)}(0) = 1$$

Substituting these values into the Taylor series gives

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

as desired.

The other options are incorrect because they do not include all powers of $x$ less than or equal to 3.