# no stationary points?

• Apr 4th 2010, 07:55 AM
osmosis786
no stationary points?
8/(2-x) - 1/(1+x)

Why would this curve have no stationary points?

The differential of this expression is : 8/(2-x)^2 + (1+x)^-2
• Apr 4th 2010, 08:22 AM
drumist
Quote:

Originally Posted by osmosis786
8/(2-x) - 1/(1+x)

Why would this curve have no stationary points?

The differential of this expression is : 8/(2-x)^2 + (1+x)^-2

If there was a stationary point, then the derivative would be zero at that point, i.e.,

$\frac{8}{(2-x)^2} + \frac{1}{(1+x)^2} = 0$

Multiplying by both denominators gives:

$8(1+x)^2 + (2-x)^2 = 0$

$8+16x+8x^2 + 4-4x+x^2=0$

$9x^2+12x+12=0$

$3x^2+4x+4=0$

Does this quadratic equation have any real solutions?
• Apr 4th 2010, 08:34 AM
osmosis786
thanks, thought of doing that but it was a 2mark question :| anyway did the discriminant and it was -288, therefore no real roots =D
• Apr 4th 2010, 12:47 PM
drumist
Quote:

Originally Posted by osmosis786
thanks, thought of doing that but it was a 2mark question :| anyway did the discriminant and it was -288, therefore no real roots =D

There is a shorter way if you are a bit more observant. Since the square of any real number must be non-negative, we can notice that the denominators of both fractions of the derivative must be positive, meaning both fractions must be positive. So we are adding two positive numbers together, which can't possibly equal zero.