Let $$f(x) = x^6 + 2x^4 + x^3 + 2x + 3$$, $$x \in R$$. Then the natural number $$n$$ for which $$\lim_{x \to 1} \frac{x^n f(1) - f(x)}{x - 1} = 44$$ is _________.

We have the polynomial function $$f(x)=x^{6}+2x^{4}+x^{3}+2x+3$$ defined for all real numbers.

First we evaluate $$f(1)$$, because it will appear repeatedly.

Substituting $$x=1$$ gives

$$f(1)=1^{6}+2\cdot1^{4}+1^{3}+2\cdot1+3 =1+2+1+2+3 =9.$$

The limit in the question is

$$\lim_{x\to1} \frac{x^{n}\,f(1)-f(x)}{x-1}.$$

Since we already know that $$f(1)=9$$, we rewrite the expression inside the limit as

$$\frac{9x^{n}-f(x)}{x-1}.$$

Notice that the form $$\displaystyle\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$$, provided the limit exists, equals the derivative of $$g(x)$$ at $$x=a$$. This is just the definition of the derivative:

$$g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}.$$

Here we may set

$$g(x)=9x^{n}-f(x).$$

Then the given limit is precisely $$g'(1)$$. Therefore we must compute the derivative of $$g(x)$$ and then evaluate it at $$x=1$$.

Differentiating term by term, we find

$$g'(x)=\frac{d}{dx}\!\bigl(9x^{n}\bigr)-\frac{d}{dx}\!\bigl(f(x)\bigr) =9n\,x^{\,n-1}-f'(x).$$

So we next need $$f'(x)$$, the derivative of the original polynomial. Differentiating each power of $$x$$ gives

$$f'(x)=\frac{d}{dx}\!\bigl(x^{6}\bigr)+\frac{d}{dx}\!\bigl(2x^{4}\bigr)+\frac{d}{dx}\!\bigl(x^{3}\bigr)+\frac{d}{dx}\!\bigl(2x\bigr)+\frac{d}{dx}\!\bigl(3\bigr).$$

Using the power rule $$\dfrac{d}{dx}(x^{k})=k\,x^{k-1}$$ we obtain

$$f'(x)=6x^{5}+8x^{3}+3x^{2}+2.$$

Now evaluate this derivative at $$x=1$$:

$$f'(1)=6\cdot1^{5}+8\cdot1^{3}+3\cdot1^{2}+2 =6+8+3+2 =19.$$

Returning to $$g'(x)=9n\,x^{\,n-1}-f'(x)$$, we substitute $$x=1$$ and the just-found value of $$f'(1)$$:

$$g'(1)=9n\cdot1^{\,n-1}-19=9n-19.$$

By the definition of $$g'(1)$$, this must equal the limit given in the problem, which is $$44$$. Hence we set

$$9n-19=44.$$

Now we solve this simple linear equation for $$n$$:

$$9n=44+19=63,$$

$$n=\frac{63}{9}=7.$$

Thus the required natural number is $$n=7$$.

Hence, the correct answer is Option 7.