- Thread starter
- #1

- Thread starter Maxers99
- Start date

- Thread starter
- #1

- Admin
- #2

- Feb 14, 2012

- 3,685

I am pretty sure you mean the domain for the second function is set for $x>0$, i.e. what we have here is the following piece-wise function:

$\displaystyle y(x)=\begin{cases}3x-2f & x\le0\\2x^2+x+5f^2 & x>0\\ \end{cases}$

Since $y$ is a continuous function, so at $x=0$, the two expressions must be equal. Can you proceed with this little hint?

- Admin
- #3

- Thread starter
- #4

- Admin
- #5

- Feb 14, 2012

- 3,685

Correct!So if x=0, it would then be -2f = 5f^2

But bear in mind that we are asked to find the real values of $f$. So now we have the quadratic equation in terms of $f$, i.e.

$-2f = 5f^2$

or

$ 5f^2+2f=0$

do you know how to solve this quadratic equation for $f$?

- Jan 29, 2012

- 1,151

At this point, either f= 0 or we can divide both sides by f.Correct!

But bear in mind that we are asked to find the real values of $f$. So now we have the quadratic equation in terms of $f$, i.e.

$-2f = 5f^2$

or

$ 5f^2+2f=0$

do you know how to solve this quadratic equation for $f$?