Finding a value to make piecewise continuous

In summary, Brian's teacher has a definition where \lim_{x\to a}f(x) = f(c) if x is within a certain range.
  • #1
BrianHare
3
0

Homework Statement



Find c such that it makes f(x) continuous.

Homework Equations



[itex]
f(x)=\begin{cases}
2x+c&x < -5\\
3x^2&-5 \leq x < 0\\
cx^2&0 \leq x\\
\end{cases}
[/itex]

The Attempt at a Solution



I know that
[tex] \lim_{x\to 5^-}3x^2 [/tex] = 2x+c
and
[tex] \lim_{x\to 0^+}3x^2 [/tex] = cx^2

Which makes the 2 points where it disconnects at (-5, 75) and (0,0)
Given that I make 2x+c = 75, where x=-5. I get C= 85, and since cx^2=0 is any real, does that mean the answer is c=85?
 
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  • #2
Welcome to PF!

BrianHare said:
I know that
[tex] \lim_{x\to 5^-}3x^2 [/tex] = 2x+c
and
[tex] \lim_{x\to 0^+}3x^2 [/tex] = cx^2

Which makes the 2 points where it disconnects at (-5, 75) and (0,0)
Given that I make 2x+c = 75, where x=-5. I get C= 85, and since cx^2=0 is any real, does that mean the answer is c=85?

Hi Brian! Welcome to PF! :smile:

Yes, c = 85.

(though you have a strange way of using lim …

you might as well say lim 3x2 = 3*(-5)2, and so on. :wink:)
 
  • #3


tiny-tim said:
Hi Brian! Welcome to PF! :smile:

Yes, c = 85.

(though you have a strange way of using lim …

you might as well say lim 3x2 = 3*(-5)2, and so on. :wink:)

My teacher has a definition where

[tex]\lim_{x\to a}f(x) [/tex] = f(c)

It was my understanding that f(x) = 3x^2, a = -5, and f(c) = 2x+c. So once I knew the answer to the limit, i knew that f(c) = 75, thus 2x+c must also be 75. Maybe I am misunderstanding the definition. Can anyone clarify?
 
  • #4
BrianHare said:
My teacher has a definition where

[tex]\lim_{x\to a}f(x) [/tex] = f(c)

:confused: that doesn't make any sense …

what is f(c) supposed to mean? :confused:

(f(c) = 2c + c or 3c2 or cc2)

does he mean [tex]\lim_{x\to a}f(x) [/tex] = f(a)?​
 

Related to Finding a value to make piecewise continuous

1. What is a piecewise continuous function?

A piecewise continuous function is a function that is defined using different equations for different intervals of the domain. This means that the function may have different behaviors or characteristics in different parts of its domain.

2. Why is it important to find a value to make a piecewise continuous function?

Finding a value to make a piecewise continuous function is important because it helps ensure that the function is continuous and has no gaps or breaks in its graph. This is necessary for many real-world applications and mathematical calculations.

3. How do you find a value to make a piecewise continuous function?

To find a value to make a piecewise continuous function, you first need to identify the different intervals or pieces of the function. Then, you can set up equations for each interval and solve for the value(s) that make the function continuous at the boundaries of the intervals.

4. Can a function be both continuous and piecewise continuous?

Yes, a function can be both continuous and piecewise continuous. A function is continuous if it has no breaks or gaps in its graph, and a piecewise continuous function is one that is defined using different equations for different intervals of the domain. Therefore, a function can be both continuous and piecewise continuous if it is defined using different equations for different intervals, but those equations are continuous within their respective intervals.

5. What are some real-world examples of piecewise continuous functions?

Some real-world examples of piecewise continuous functions include a piecewise defined cost function for a business, where the cost may change at different levels of production, and a piecewise defined velocity function for a moving object, where the velocity may change depending on the terrain or obstacles encountered on its path.

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