<s>[INST] How to solve $f(-4)$ if $f(x)=\frac{x^3+64}{x+4}$? I need to solve $f(-4)$ if $f(x)=\frac{x^3+64}{x+4}$. I have changed the form of the function, but for some reason I'm still not getting the right answer.
My Steps:
$$\frac{x^3+64}{x+4}$$
apply $(a+b)^3=(a+b)(a^2+2ab+b^2)$ formula
$$=\frac{(x+4)(x^2+8x+16)}{x+4}$$
$$=x^2+8x+16$$
Insert $-4$ for $x$.
$$(-4)^2+8(-4)+16 = 16-32+16 = 0$$
 [/INST] There is no value $f(-4)$ since $-4$ is not in the domain of the function by mere fact of the division by zero that results. Even if the expression simplifies, it doesn't change the removable discontinuity being present.
In any event, if you want to find an almost-equivalent expression for $f$ that gets rid of the discontinuity, note that you factor a sum of two cubes as so:
$$a^3 + b^3= (a+b)(a^2 - ab + b^2)$$
(You made a sign error, thus why I noted it. In general, you might see this written as $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ to account for the sign change.)
Thus, $x^3 + 64 = x^3 + 4^3$ factors as
$$x^3 + 64 = (x+4)(x^2 - 4x + 16)$$
Then dividing by $(x+4)$ gives you $g(x) = x^2 - 4x + 16$. In this sense, you have $g(-4)$ defined.
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