What is meant by steepest ascent I am trying to understand the following from my notes, in relation to Newton Like methods.
if 
$x^{(t+1)}=x^{(t)}-(M^{(t)})^{-1}g'(x^{(t)})$
is not guarnteed to be uphill.
However if $M^{(t)}=-I$ (the identity), then 
$x^{(t+1)}=x^{(t)}+g'(x^{(t)})$
assures steepest ascent and is uphill.
How? What if $g'(x^{(t)})$ is negative?
 A: Note: Your question poses this problem using some notation that suggests that $g$ is a univariate function, and some notation that suggest that $g$ is a multivariate function.  I am going to pose my answer for the more general case where $g$ can be multivariate, so I will use the notation $\nabla g$ for the gradient of $g$, and I will only use $g'$ for the special case where $g$ is univariate.

There is an important result in calculus which tells us that if $g$ is a differentiable function (whether a univariate or multivariate function) then the gradient vector is orthogonal to the level sets, and in particular, the function increases fastest in the direction of its gradient, and decreases fastest in the direction of the negative of its gradient.  So if you are presently at the point $x^{(t)}$ and you want to go in the direction of the steepest ascent, then you would go in the direction of the gradient vector (usually by some small distance $\gamma$) to the new point:
$$x^{(t+1)} = x^{(t)} + \gamma \nabla g (x^{(t)}).$$
Now, in the equations you present it looks like you have set the step size to $\gamma = 1$, so you get:
$$x^{(t+1)} = x^{(t)} + \nabla g (x^{(t)}).$$
In your question you wonder what would happen if $g'(x^{(t)})$ is negative.  (Presumably you are now treating $g$ as univariate.)  That wouldn't matter.  All it would mean is that the function is decreasing and therefore the steepest ascent is to go backwards on the horizontal axis ---i.e., move to some $x^{(t+1)} < x^{(t)}$.  If your step size is sufficiently small, this should generally result in an ascent of the function output.  If you draw a plot of a smooth univariate function and pick a point with negative slope you will see that ascending the function requires you to go backwards on the horizontal axis.
