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lisabon 2012 [21]
3 years ago
14

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

Mathematics
1 answer:
9966 [12]3 years ago
4 0

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

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