# One-one function

• May 30th 2011, 02:54 AM
Punch
One-one function
Find the derivative of $h(x)$ where $h(x)=\frac{1}{f(x)}$ and $f(x)=tanx+1$, $0

Hence show that $h(x)$ is a one-one function

I have found the derivative of h(x) to be $-\frac{sec^2x}{(tanx+1)^2}$ but am confused on how i could use this to show $h(x)$ is one-one
• May 30th 2011, 03:08 AM
HallsofIvy
If a function is either always increasing or always decreasing then it is one to one,
• May 30th 2011, 03:23 AM
Punch
oh... so u mean that the gradient is always decreasing or increasing, then it is one to one?
• May 30th 2011, 04:01 AM
Punch
Quote:

Originally Posted by HallsofIvy
If a function is either always increasing or always decreasing then it is one to one,

sorry but i didnt quite get what you meant
• Jun 4th 2011, 05:11 PM
HallsofIvy
Do you understand what "one to one" means? A function, f, is one to one if and only if f(x)= f(y) implies x= y. If f is increasing, we can use "proof by contradiction": Suppose f(x)= f(y) with $x\ne y$. Then either x> y so that f(x)> f(y) or x< y f(x)< f(y), either contradicting f(x)= f(y). The same argument works if f is decreasing.
• Jun 4th 2011, 05:13 PM
HallsofIvy
Quote:

Originally Posted by Punch
oh... so u mean that the gradient is always decreasing or increasing, then it is one to one?

No, I mean that if the function is increasing or decreasing (it gradient (derivative) is non-negative or non-positive) then it is one to one.
• Jun 4th 2011, 05:25 PM
Ackbeet
Quote:

Originally Posted by HallsofIvy
No, I mean that if the function is increasing or decreasing (it gradient (derivative) is non-negative or non-positive) then it is one to one.

Actually, to be technically precise, the derivative would have to be positive or negative; it could only be zero at isolated points and not on any interval.
• Jun 4th 2011, 08:49 PM
Corpsecreate
For a function to be one-to-one, the function must pass the vertical line test and the horizontal line test for all values of x.