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Find a real number for a continuous function

Maxers99

New member
Feb 26, 2014
2
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
Feb 14, 2012
3,685
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Feb 26, 2014
2
So if x=0, it would then be -2f = 5f^2
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
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
Jan 29, 2012
1,151
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 $f$?