Hey guys this is my question below
Let $$f:[a,b]\to \mathbb R$$ be continuous on [a,b] and twice differentiable on (a,b) with $$f'(x) \neq 0 , f"(x) = 0, \forall x \in (a,b) $$
Suppose that $$\frac{f(x)}{f'(x)} \in [-(b-a),0], \forall x \in (a,b)$$
and there exists a constant k s.t $$\frac{f(x)f"(x)}{(f'(x))^2} \leq k < 1, \forall x \in (a,b)$$
Use the fixed point theorem to prove that for any $$x0 \in [a,b]$$
Newton's iteration converges to the root of f in [a,b]
My attempt
This is what I know a fixed point is where $$x = g(x)$$ for a given function therefore the x value is the same as the functional value hence the name fixed point.
Therefore given a root of the equation p this will give the follwoing
$$g(p) = p$$
I know that there is a theorem that states the following if
$$ f \in C[a,b] and f(x) \in [a,b], \forall x \in [a,b]$$
then f has a fixed point in [a,b]
In addition if $$f'(x)$$ exists on (a,b) and there exists a positive constant k < 1 s.t.
$$|f'(x)| \leq k < 1$$ then the fixed point is unique.
Therefore my thinking is for newtons method given by the followng
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
$$\implies$$
$$x_{n+1} = g(x_n)$$
Therefoe since f is continuous from the question
$$a,b > 0$$
$$b> a$$
based on the condition given $$|\frac{f(x)f"(x)}{[f'(x)^2]}| \leq k < 1$$
the function converges to unique fixed point as
$$\frac{|f(x)f"(x)|}{|f'(x)^2)|} = 0$$
As well as
$$\frac{f(x)}{f'(x)} \in [-(b-a),0]$$
$$\implies$$
$$f(x) > 0 $$
$$f'(x) < 0$$
Therefore since the derivative of the function is < 1 the fixed point is unique and converges to the root of f in [a,b] based on both conditions but im not sure if this would suffice to prove what is required. Can anyone tell me if this is correct and if not the method by which i prove it.