I know points of increase and decrease can be found using the derivative. To take an example of $f(x) = x^2$, we have $\frac{d}{dx}f(x) = 2x$. The slope is constant. When $\frac{d}{dx}f(x) = 0$, we have a critical point, $c=0$.

But how do I determine the intervals of increase/decrease given this info? Thus far I've been testing $f(c-0.1)$ and $f(c+0.1)$ and comparing the values. Is there another method to determine the intervals of increase/decrease?

It becomes labor intensive when I have a curve with several inflection points and need to find all such intervals.

  • $\begingroup$ What do you mean? Wherever the derivative is positive, the map increases. Where it’s negative, the map decreases. $\endgroup$
    – k.stm
    Jun 8, 2019 at 14:49
  • 1
    $\begingroup$ You don't have to do $c-0.1$. You can pick any point in the intervals between critical points and check the sign of the derivative, including one that is easy to compute. $\endgroup$ Jun 8, 2019 at 15:11

1 Answer 1


As you guessed, we should be looking at the derivative. Why? Because by definition, derivative at a point is the slope of the tangent line to the function at that point. Thus, if $f^{\prime}(a) > 0$ then the tangent line has a positive slope, hence the function must be increasing. Similarly, if $f^{\prime}(a) < 0$ then the tangent line has negative slope, so the function must be decreasing at this point.

Now the key ingredient is for the function to change from increasing to decreasing, it must create a "peak", or a "valley". But from the perspective of derivative, at the peak or the valley, the derivative must either be equal to $0$ or "undefined" (since the peak/valley can be really sharp! hence not differentiable).

Knowing this fact, we then can use derivative to split the domain of the function into many intervals, where the splitting occurs at the points where the derivative is $0$ or undefined. However, just because the derivative is $0$ or undefined, it doesn't mean the function will changes "sign" (increasing to decreasing or decreasing to increasing)... for example: look at $f(x) = x^3$ . But if the function changes sign, it must happen at these points.

So for your problem, you have $f(x) = x^2$. Hence, $f^{\prime}(x) = 2x $.

$f^{\prime}(x) = 2x = 0 \ \ \ \Rightarrow \ x= 0$

and $f^{\prime}(x)$ will never be undefined so we don't have to worry about that.

So the splitting now will occurs at $0$ as you suggested. This mean if the function changes its "increasing/decreasing" behavior, it must happen at this point.

To check this, we plug in any point to the left of $0$ and the right of $0$ into $f^{\prime}(x)$.

For the left of $0$, let's pick $x = -1 $, then $f^{\prime}(-1) = 2(-1) = -2$. DECREASING - since the derivative is negative!

For the right of $0$, let's pick $x = 1 $, then $f^{\prime}(1) = 2(1) = 2$. INCREASING - since the derivative is positive!

Thus, you can see that the function $f(x)$ is decreasing on the left of $0$, and increasing on the right of $0$. Therefore, geometrically, you can see that the function must have a "Valley" (minimum) at $x=0$.

Hope this help.


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