Again find a such f is continuous at x = 2

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In summary, the conversation discusses finding a value for a that makes the function f(x) continuous at x = 2. The solution involves setting the two limits (as x approaches 2 from the left and right) equal to each other and solving for a. The answer of a = 3 works, but it is important to be careful when cancelling variables and to check the limits to verify the solution.
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Homework Statement



Another similar question. I will appreciate any input!

[tex]\[f(x) = \left\{\begin{matrix}
x^{2}+a^{2} & x \leq 2\\ 2x+3a
& x > 2
\end{matrix}\right.\][/tex]

find a such f is continuous at x = 2

Homework Equations



Well... the definition of continuity


The Attempt at a Solution




So I have the first condition limit as x approaches 2 gives 4 + a^2
The second limit should equal to the first limit to agree (both left and right) and that limit gives 4+3a.
Equate the two, I solve for a

4 + 3a = 4+a^2
3a = a^2
a = 3

Am I right?

Thank you !
 
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  • #2
So are the limits the same then, once you plug 3 in for a?
Also, are you sure 3 is the only a that works? Are there any others?
 
  • #3
Hi, thank you for the inputs.

So are the limits the same then, once you plug 3 in for a?
Well if I started out with limit as x approaches 2 x^2 + a^2 without knowing what a is, then I should end up with (2)^+a^2.
And equate that with the second limit I would end up with a^2 = 3a which cancels 1 a on the left side, and 3 = a.

I am sure if a = 0, then you will have the same limit too.
But if this is such case, then something isn't right in my calculation (since you can't get 2 distinct solutions from a single calculation - which reduces to only a = 3).
 
  • #4
One must always be careful when cancelling variables (as soon as you cancel, you assume your variable cannot be equal to 0). When solving for a in a^2 = 3a, you can look at it as the quadratic equation a^2-3a = 0. Then solve this using your favorite technique to solve for the roots of a quadratic (quadratic formula, completing the square, factoring etc). Note this one is particularly simple if you just factor out an a.
Whatever you get from these values of a, plug them back into your equations and then take the limits. If they are the same thing, then yay, that value of a works!

As for saying that you can't get 2 distinct solutions from a single calculation, this is not true at all. It entirely depends on the calculation. For example, [tex]\sqrt4[\tex] is one calculation, but gives two answers: 2 and -2

Your answer of a=3 works, and you should verify this by calculating the limits (as x goes to 2) of x^2-3^2
and 2x-3(3), and seeing that they are the same.
 
  • #5
Oh right. a(a - 3) gives 0 and 3 respectively.

Thank you.

PS: Yeah I was referring to a = 3 cannot produce 2 distinct solutions, but yes certainly i should be careful with the quadratic.
 

Related to Again find a such f is continuous at x = 2

1. What does it mean for a function to be continuous at a specific point?

When a function is continuous at a specific point, it means that the limit of the function at that point exists and is equal to the value of the function at that point. In simpler terms, there are no breaks or jumps in the graph of the function at that point.

2. How can you determine if a function is continuous at x = 2?

To determine if a function is continuous at x = 2, you need to check if the limit of the function as x approaches 2 exists and is equal to the value of the function at x = 2. In other words, you need to evaluate the left-hand limit, the right-hand limit, and the value of the function at x = 2, and make sure they are all equal.

3. Can a function be continuous at x = 2 if the limit does not exist?

No, a function cannot be continuous at a specific point if the limit does not exist at that point. This means that there is a break or a jump in the graph of the function at that point, which violates the definition of continuity.

4. What are the conditions for a function to be continuous at a specific point?

For a function to be continuous at a specific point, three conditions must be met: 1) the limit of the function at that point must exist, 2) the function must be defined at that point, and 3) the limit must be equal to the value of the function at that point.

5. How is continuity important in mathematics and science?

Continuity is an important concept in mathematics and science because it allows for the smooth and consistent behavior of functions. It also helps in understanding the behavior of real-world phenomena, such as motion and growth, which can be modeled using continuous functions. Additionally, many mathematical theorems and principles rely on the assumption of continuity.

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