1. ## Continuous functions

If possible, choose k so that the following function is continuous on any interval.

f(x)= (4x^(3)-8x^(2))/(x-2) for x not= 2

k for x=2

alright, i need help on how to start this one, any suggestions would be very helpful, thank you

2. $\frac{{4x^3 - 8x^2 }}{{x - 2}} = 4x^2 ,\quad x \ne 2$

3. Originally Posted by Plato
$\frac{{4x^3 - 8x^2 }}{{x - 2}} = 4x^2 ,\quad x \ne 2$

great, so what exactly do you do to find that it = 4x^2?

4. You cannot find some value k such that $f(2)=k$ is continuous with the rest of the function.

5. Originally Posted by mathlete
great, so what exactly do you do to find that it = 4x^2?
Do you know what it means for the function to be continuous at x=2?

6. Originally Posted by Plato
Do you know what it means for the function to be continuous at x=2?
oh, ok i got it...duh, thanks alot

7. Originally Posted by mathlete
If possible, choose k so that the following function is continuous on any interval.

f(x)= (4x^(3)-8x^(2))/(x-2) for x not= 2

k for x=2

alright, i need help on how to start this one, any suggestions would be very helpful, thank you

$\frac{4x^3-8x^2}{x-2} =4x^2 \frac{x-2}{x-2}$

so when $x \ne 2$:

$\frac{4x^3-8x^2}{x-2} =4x^2$

Hence:

$\lim_{x \to 2} \frac{4x^3-8x^2}{x-2} =4\times 2^2=16$

So if you make $k=16$ then $f$ will be continuous.

RonL