Finding k for Continuous Function on Any Interval

[Tom McCurdy]

Alright this is confusing me a bit...

Find k so that the following function is continuous on any interval?
$$f(x)=kx$$ if $$0 \leq x \leq 2$$ and $$f(x)=5x^2$$ if $$2\leq x$$

Alright I know the answer is 10, but I don't understand how you get there

I mean I just doubled 5, because I took it off an example in the book that had answer

but i would like be able to do eveutnally do

If possible choose k so that the following function is continusous on any function

$$f(z) = \left\{ \begin{array}{rcl} \frac{5x^3-10x^2}{x-2} & \mbox{ } & x\neq2 \\ k & \mbox{ } & x=2 \end{array}\right.$$

 

[Tzar]

Alright this is confusing me a bit...

Find k so that the following function is continuous on any interval?
$$f(x)=kx$$ if $$0 \leq x \leq 2$$ and $$f(x)=5x^2$$ if $$2\leq x$$

You want the function to be continuous at x=2. Therefore, k*2 = 5*2^2. Solve for k

 

[Tom McCurdy]

So would tthe answer to the second one be false

?

 

[Tzar]

So would tthe answer to the second one be false

?

No the answer is true if you make k=20.

You see, you can factor out 5x^2 from the top, and then the (x-2) will cancel. If you then sub in x=2 you get 20. Thus if you make k=20, the function becomes continous.