f '(x)=x² - 2 + 1/x²
so for increasing function, f '(x) > 0
⇒ x² - 2 + 1/x² > 0
so my question is how do i go about from here to show this is true for all real values of x.
the answer provided is the equation simplified to (x - 1/x)² > 0
my other question would be how did they simplify that eqn?
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- Thread Starter
- 12-04-2006 20:44
- 12-04-2006 20:54
x² - 2 + 1/x² = x2 - x/x - x/x + 1/x2 = (x - 1/x)(x - 1/x) = (x - 1/x)2Last edited by CalculusMan; 12-04-2006 at 20:56.
- 12-04-2006 20:57
So you want to prove f(x) is an increasing function?
The condition is that the derivative of f(x) is greater than zero for all x.
It provides you with the expression for the derivative: x2 - 2 + 1/x2
This can be FACTORISED to (x - 1/x)2. If it's not obvious why, look at the following:
x2 - 2 + 1/x2
Take out 1/x2 as a factor:
x-2(x4 - 2x2 + 1)
Let y = x2
y-1(y2 - 2y + 1)
There's a familar quadratic in the brackets, which you know how to factorise...
Then put back in y = x2...
x-2(x2 - 1)2
Now, x-2 = (x-1)2, so you can take it inside the bracket as x-1...
(x - x-1)2
Which is equivalent to (x - 1/x)2.
It's just a simple factorisation. If you expand that bracket, you'll get the original expression.
The final step of the argument is to say that since the derivative is expressed as a perfect square, then there is no value of x which will make it negative. Also, since there is a 1/x term, x must not be zero!
Hence, the derivative must always be greater than zero.
- Thread Starter
- 13-04-2006 01:01
thank you worzo for the clear explanation. +ve rep has been sent!!