• May 16th 2010, 03:23 AM
cistudent
List the values at which the function

$

f(x) = \frac{1}{x^2+9}
$

Is not defined.

This question does not have an answer in my text book so I cannot deduce where I have gone wrong, though actually I haven't gotten anywhere...just staring at it and repeating it to myself. I've tried factorising it out, but that doesn't work see (x+3)(x-3)..... ${x^2}$ must be a positive number right?
• May 16th 2010, 03:56 AM
Tekken
well you did the right thing in factorising the bottom line. The function is not defined if the denominator is 0

=> f(x) not defined at $x^{2} + 9 = 0$

=> not defined $x = 3$ and $x = -3$
• May 16th 2010, 04:01 AM
e^(i*pi)
Quote:

Originally Posted by Tekken
well you did the right thing in factorising the bottom line. The function is not defined if the denominator is 0

=> f(x) not defined at $x^{2} + 9 = 0$

=> not defined $x = 3$ and $x = -3$

$f(3) = \frac{1}{3^2+9} = \frac{1}{18} \text{ and } f(-3) = \frac{1}{(-3)^2+9} = \frac{1}{18}$ which is clearly defined.

$(x+3)(x-3) = x^2-9$. Note that $(-3)^2 = 3^2 = 9$ and hence $x^2+9 \geq 9$ over the real numbers.

This means there are no real values of x for which f(x) is undefined.

There are complex ones though: $3i$ and it's conjugate $-3i$. This comes from solving $x^2+9 =0$ over the complex numbers.

Rewrite f(x) in terms of i (where $i^2 = -1$)

$x^2 + 9 = x^2 - (3i)^2$

The difference of two squares can now be used

$x^2-(3i)^2 = (x-3i)(x+3i) = 0$
• May 16th 2010, 04:18 AM
cistudent
Ah so a trick question. I always fall for these. I am so mathematically challenged