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BasiC MatH

  • #1
1,010
1
Alright this is confusing me a bit...

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

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

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

Answers and Replies

  • #2
74
0
Tom McCurdy said:
Alright this is confusing me a bit...

Find k so that the following function is continuous on any interval?
[tex] f(x)=kx [/tex] if [tex]0 \leq x \leq 2 [/tex] and [tex]f(x)=5x^2 [/tex] if [tex] 2\leq x [/tex]
You want the function to be continuous at x=2. Therefore, k*2 = 5*2^2. Solve for k
 
  • #3
1,010
1
So would tthe answer to the second one be false

?
 
  • #4
74
0
Tom McCurdy said:
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.
 

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