Find sufficient and necessary conditions in which the function has a unique fixed point Let us consider the function 
$$f(x)=αx^{θ}+β$$ where $α>0,β<0,θ>0$.
My question is: Find sufficient and necessary conditions in which this function has a unique fixed point in the interval $[0,1]$.
 A: (Quoting @hardmath as well since he commented first about a similar way of solving it)
Recall that a fixed point means : 
$$f(x_0) = x_0 \Rightarrow ax_0^\theta + \beta = x_0 \Leftrightarrow ax_0^\theta+\beta - x_0 = 0 \Leftrightarrow x_0(ax_0^{\theta - 1}-1) + \beta=0$$
Thus all you need to do, is determine when the function :
$$g(x) = x(ax^{\theta - 1}-1) + \beta $$
has a unique solution in the interval $[0,1]$.
The values of $g(x)$ at the end points of the given interval, are : 
$$g(0) = \beta$$
$$g(1) = a-1+\beta$$
For the function $g(x)$ to have at least one root in the interval $[0,1]$, one can apply Bolzano's Theorem since the function is continuous in it. Then, if : 
$$g(0) \cdot g(1) < 0 \Rightarrow (a-1+\beta)\beta < 0$$
the function $g(x)$ will have at least one root in $(0,1)$. Since $\beta <0$ then you'll need : 
$$a-1+\beta > 0 $$
For this root to be unique, it would be enough if you could show that $g(x)$ is strictly monotonic. Studying the derivative of $g(x)$, we'll have : 
$$g'(x) =a\theta x^{\theta-1}-1$$
To be strictly monotonic, you want one of the following, since strictly monotonic means increasing or decreasing in the given interval : 
$$g'(x) > 0 \Rightarrow a\theta x^{\theta -1} -1>0 \quad \forall x\in (0,1)$$
$$g'(x) < 0 \Rightarrow a\theta x^{\theta -1} -1<0 \quad \forall x\in (0,1)$$
Joining these $2$ with the first case in $2$ different cases, you'll have sufficient conditions for strictly one root of $g(x)$ in $[0,1]$ or in other words a unique fixed point of $f(x)$, if you can give solutions to the following systems of inequalities for all $x \in (0,1)$ : 
$$\begin{cases} a-1+\beta < 0  \\ a\theta x^{\theta -1} -1>0 \end{cases}$$
$$\text{or}$$
$$\begin{cases} a-1+\beta < 0  \\ a\theta x^{\theta -1} -1<0 \end{cases}$$
