If y' is positive at a value of x, then y is increasing there. Is the converse true?

Question

I know it's a dumb question, but I really don't know it. I've thought of the derivative of y = x but it's still positive and the book says the answer is no.

Answer 1

The question deals with a function $y=f(x) $ if $x$ and it's behavior at a specific point $x$.

It is better to use a different symbol $c$ to represent the specific point. So let the derivative $y'=f'(x) $ at $x=c$ be positive ie $f'(c) >0$. Since $$f'(c) =\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}$$ it follows that the fraction under above limit is positive for all sufficiently small values of $h$. This means that $f(c+h) - f(c) $ has same sign as that of $h$. Let's take $h>0$ then it means that $f(c+h) >f(c) $. If $h<0$, say $h=-k,k>0$, then $f(c+h) =f(c-k) <f(c) $.

The above conclusion can be worded in the following manner. If $f'(c) >0$ then there is an open interval $I$ with $c\in I$ such that if $x\in I, x<c$ then $f(x) < f(c) $ and if $x\in I, x>c$ then $f(x) >f(c) $. This behavior is expressed more concisely as $f$ is strictly increasing at point $c$. It does not mean that $f$ is strictly increasing in some interval like $I$ containing $c$.

On the other hand the converse does not hold true. In other words if $f$ is strictly increasing at point $c$ and $f'(c) $ exists then it does not necessarily imply that $f'(c) >0$. One can only conclude $f'(c) \geq 0$. An example is $f(x) =x^3$ and $c=0$. Convince yourself that $f$ is strictly increasing at $0$ but $f'(0)=0$.

Answer 2

Yes, the converse is true, at least if your function is sufficiently "nice" (i.e. differentiable). The derivative is just the gradient of the function, so if the derivative is positive, then so is the gradient, and the function increases. But if the derivative doesn't exist (i.e. the function is not differentiable), then even if it seems to be increasing on some interval based on, say, its graph, its derivative would not be positive because it doesn't exist. For example, look at the Weierstrass function.

Answer 3

Have a look at $f:\mathbb R\to\mathbb R, x\longmapsto x^3$.

$f$ is strictly increasing, namely $f(x_0)<f(x_1)$ whenever $x_0<x_1$, and this even at the origin,
but $f'(0)$ is ...

Answer 4

First a monotonic function is derivable almost everywhere. So if you add a continuity hypothesis around $x_0$ then you'll get the derivability in $x_0$.

From there $f'(x_0)=\lim\limits_{x\to x_0}\underbrace{\dfrac{f(x)-f(x_0)}{x-x_0}}_{>0\text{ by strict increasing}}\ge 0$

So even if you consider strict monotonicity, the inequality will become loose by limit operation.

So the converse may not be true:

• because the function is not continuous in $x_0$, e.g. $f(x)=\lfloor x\rfloor$, it is increasing but $f'(1)$ is not defined for instance since it has different values on the left and on the right of $x_0=1$.
• because the derivative is zero and not strictly positive, e.g $f(x)=\operatorname{sgn}(x)\ e^{-1/x^2}$ whose all n-derivatives in $0$ are zero.

Anyway, if you consider the function locally continuous around $x_0$ and positive in a loose sense, then the converse is true.

Answer 5

Yes, the converse is true. In geometric terms, $f'(x)$ denotes the slope of $f(x)$ at $x$. If the slope is positive, then the function is increasing. Note that $f'(x)$ can be decreasing, but still positive. This means that $f(x)$ is still increasing but is what we call concave down: that is, increasing more and more slowly.