Find the constant value 'a' such that the fcn is continuous

[punkyfish0725]

Consider the following:

g(x) = {x^2 - a^2. . . . . .x does not equal a
. . . . . .{. . .9. . . . . . . . . .x = a

Find the constant a such that the function is continuous on the entire real line.

I do not understand this question. Please help. Thank you!
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Edited by stapel -- Reason for edit: restoring spacing

 

[pka]

Are you sure that you have written this correctly?
I can prove that \(\displaystyle \lim _{x \to a} \left( {x^2 - a^2 } \right) = 0 \not= 9\) for every value of a.
Hence, no matter what value we assign a, g will not be continuous at a.

OR

Could you mean \(\displaystyle \frac{{\left( {x^2 - a^2 } \right)}}{9},\quad x \not= a???\)

 

[punkyfish0725]

i have it written exactly as it says , it's definitely not the second bit you wrote

 

[stapel]

i have it written exactly as it says , it's definitely not the second bit you wrote

Then the previous reply is correct: There is no solution.

For x = a, you have x² - a² = a² - a² = 0, regardless of the value of a.

You've studied limits (back at the beginning of calculus). There is no way for something that is approaching zero (that is, something whose limit, as x approaches a, is zero) to be continuous if its value at x = a is anything other than zero. If g(a) = 9, then there is no way for this to be made continous. Sorry.

Eliz.