MHB Find a real number for a continuous function

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To find a real number f for the continuous piecewise function, the two expressions must be equal at x=0. This leads to the equation -2f = 5f^2, which simplifies to the quadratic equation 5f^2 + 2f = 0. The solutions to this equation are f = 0 or f = -2/5. Thus, the real values of f that ensure the function is continuous are 0 and -2/5. The discussion emphasizes the importance of continuity at the boundary of the piecewise function.
Maxers99
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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 }
 
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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?
 
At the risk of being somewhat redundant, to put what anemone has stated in the parlance of limits, we require:

$$\lim_{x\to0^{-}}(3x-2f)=\lim_{x\to0^{+}}\left(2x^2+x+5f^2 \right)$$
 
So if x=0, it would then be -2f = 5f^2
 
Maxers99 said:
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 $f$?
 
anemone said:
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 $f$?
 
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