Question

Find the value of f(2), so that the function $f\left(x\right)=\frac{12x-24}{{(4+2x)}^{1/3}-2},x\ne 2$ is continuous everywhere

Answer 1

For f to be continuous at $x=2$, we must have,
$f\left(2\right)=\underset{x\to 2}{lim}\frac{12x-24}{{(4+2x)}^{1/3}-2}$

Put $a=(4+2x{)}^{1/3},b=2\Rightarrow a^3-b^3=(a-b\left)\right(a^2+ab+b^2)$

$f\left(2\right)=\underset{x\to 2}{lim}\frac{12(x-2)}{(4+2x-2^3)}\left[\right(4+2x{)}^{2/3}+(4+2x{)}^{1/3}.2+2^2]$

$=\underset{x\to 2}{lim}\frac{12(x-2)}{2(x-2)}\left[\right(4+2x{)}^{2/3}+(4+2x{)}^{1/3}.2+2^2]$

$=\frac{12}{2}.12=72$