Again find a such f is continuous at x = 2

[999iscool]

Homework Statement

Another similar question. I will appreciate any input!

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

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 !

 

[rakalakalili]

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?

 

[999iscool]

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).

 

[rakalakalili]

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, \sqrt4[\tex] is one calculation, but gives two answers: 2 and -2<br /> <br /> 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<br /> and 2x-3(3), and seeing that they are the same.

 

[999iscool]

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.