Slope at point X and the next closest value to X

[Atran]

Hi,

Say we have f(x) = sin(x), then f'(x) = cos(x) and f''(x) = -sin(x)

Let x = π/2, the slope at that point is zero: cos(x) = 0. The second derivative gives, -sin(x) = -1: that means the slope of the first derivative at point x is negative. Thus for the next closest value to x, the slope of the original function is negative.

I want to know why the next closest value, say t, to any x has f(t)>f(x) if the slope at point x is positive, and f(t)<f(x) if the slope at point x is negative?

Thanks for help.

 

[HallsofIvy]

You will have to explain what you mean by "next closest point".

Since you say nothing about the domain of your fuction, the default is the real numbers. And, given a real number, x, there is no "next closest point".

 

[Atran]

Sorry for that. I mean a value bigger than x, or to the right of x.

 

[Stephen Tashi]

Say we have that means the slope of the first derivative at point x is negative. Thus for the next closest value to x, the slope of the original function is negative.<br /> <br /> .

<br /> <br /> The second derivative of f(x) gives the slope of the derivative of f(x), not the slope of f(x) itself. For example, in a scenario from physics, if you throw a ball straight up, then at the time t when it stops going up and starts to fall back down, the velocity is 0. But the acceleration (due to gravity) is negative throughout the whole time that the ball is in motion.

 

[HallsofIvy]

Sorry for that. I mean a value bigger than x, or to the right of x.

That doesn't answer my question. There exist an infinite collection of numbers "bigger than x". what do you mean by "the next closest point"?

 

[WannabeNewton]

I want to know why the next closest value, say t, to any x has f(t)>f(x) if the slope at point x is positive, and f(t)<f(x) if the slope at point x is negative?

Hi there! In essence, the second derivative encodes the curvature of the graph of a function. Let $f:A\subseteq \mathbb{R}\rightarrow \mathbb{R}$ be a function; recall that $a\in A$ is a local minimum (maximum) for $f$ if there exists a $\delta > 0$ such that $a$ is a minimum (maximum) on $(a - \delta,a+\delta)\cap A$. Recall that this implies, for $f$ differentiable at $a$, that $f'(a) = 0$; we say such points are stationary points.

As we know the converse need not hold in general but there is a theorem that will tell us the behavior of the graph of $f$ around stationary points depending on how the second derivative behaves there; it is this that we are interested in. In particular, suppose there exists $a\in A$ such that both $f'(a)$ and $f''(a)$ exist and $f'(a) = 0$. If $f''(a) > 0$ then $f$ has a local minimum at $a$ and if $f''(a) < 0$ then $f$ has a local maximum at $a$.

So what does this mean geometrically? Well let's first assume that $f''(a) > 0$. We then know that there exists, by definition of a local minimum, a neighborhood $(a - \delta,a+\delta)\cap A$ of $a$, in $A$, on which $a$ is a minimum, meaning $f(a)\leq f(x)$ for all $x\in (a - \delta,a+\delta)\cap A$. If you picture this through a graph what you will see is that $(a,f(a))$ is the trough of the graph amongst the nearby points on the graph associated with the aforementioned neighborhood. So it is in effect bending "outwards" near $(a,f(a))$. Similarly, for $f''(a) < 0$ (local maximum), the graph will bend "inwards" near $(a,f(a))$.