1. ## One-one function

Find the derivative of $\displaystyle h(x)$ where $\displaystyle h(x)=\frac{1}{f(x)}$ and $\displaystyle f(x)=tanx+1$, $\displaystyle 0<x<\frac{\pi}{2}$

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

I have found the derivative of h(x) to be $\displaystyle -\frac{sec^2x}{(tanx+1)^2}$ but am confused on how i could use this to show $\displaystyle h(x)$ is one-one

2. If a function is either always increasing or always decreasing then it is one to one,

3. oh... so u mean that the gradient is always decreasing or increasing, then it is one to one?

4. 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

5. 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 $\displaystyle x\ne y$. 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.

6. 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.

7. 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.

8. 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.