Let \(f(x) = \begin{cases} 5x^2+2&\text{if } x\le a, \\ 11x &\text{if } x>a. \end{cases} \)Find the smallest possible value for \(a\) if the graph of \(y=f(x)\) is continuous (which means the graph can be drawn without lifting your pencil from the paper).

Guest Jun 13, 2018

#1**0 **

I'm not sure how to do this, but from your definition what continuous is, I could attempt it.

Set the two functions equal to each other to find where they intersect, so you can draw them w/o lifting your pencil once you reach that point(s).

5x^{2}+2=11x

5x^{2}-11x+2=0

x=2, x=1/5

The smallest value of x where they intersect is 1/5, so a=1/5.

Graphed it and it seems to work so probably a=1/5.

ChowMein
Jun 13, 2018