To find the Taylor polynomial of degree 33 approximating the function $f(x) = x - 2x - 1$ near $x = 0$, let's first clarify the function and its derivatives.

It looks like the given function is written ambiguously. I interpret the function as:

$f(x) = x - 2x - 1 = -x - 1$

This simplifies to a linear function:

$f(x) = -x - 1$

Taylor Polynomial of Degree 33:

The Taylor series for a function $f(x)$ about $x = 0$ is given by:

$T_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n$

Since the function $f(x) = -x - 1$ is linear, its higher-order derivatives (from the second derivative onward) will be zero.

1. $f(x) = -x - 1$
2. First derivative: $f'(x) = -1$
3. Second derivative: $f''(x) = 0$
4. All higher-order derivatives: $f^{(n)}(x) = 0$ for $n \geq 2$

Now, we compute the Taylor polynomial:

1. $f(0) = -(0) - 1 = -1$
2. $f'(0) = -1$

Since all higher-order derivatives are zero, the Taylor polynomial of degree 33 (or any degree greater than 1) will just be a linear approximation:

$T_{33}(x) = -1 - x$

Thus, the Taylor polynomial of degree 33 for $f(x) = -x - 1$ near $x = 0$ is:

$T_{33}(x) = -x - 1$

Would you like more details or further clarification?

Related Questions:

1. How do you compute higher-order Taylor polynomials for more complex functions?
2. What is the general formula for a Taylor series expansion?
3. How does the behavior of a linear function affect its Taylor series?
4. What happens when you approximate a quadratic or cubic function with a Taylor series?
5. Can we use Taylor polynomials to approximate functions like $\sin(x)$ or $e^x$?

Tip:

For non-linear functions, the Taylor polynomial's accuracy improves with higher degrees, but for a linear function like $-x - 1$, it only requires a first-degree polynomial.