# Find a real number for a continuous function

#### Maxers99

##### New member
How would I go about doing this?

Find a real number f so that: is a continuous function

y = { 3x - 2f if x is less than or equal to 0. }
{ 2x2 + x + 5f2 if x is less than 0 }

#### anemone

##### MHB POTW Director
Staff member
Hi Maxers99, welcome to MHB!

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?

#### MarkFL

Staff member
At the risk of being somewhat redundant, to put what anemone has stated in the parlance of limits, we require:

$$\displaystyle \lim_{x\to0^{-}}(3x-2f)=\lim_{x\to0^{+}}\left(2x^2+x+5f^2 \right)$$

#### Maxers99

##### New member
So if x=0, it would then be -2f = 5f^2

#### anemone

##### MHB POTW Director
Staff member
So if x=0, it would then be -2f = 5f^2
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 $f$?

#### HallsofIvy

##### Well-known member
MHB Math Helper
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.

$-2f = 5f^2$
At this point, either f= 0 or we can divide both sides by f.

or

$5f^2+2f=0$

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