The sign of the first derivative tells you whether a function is rising or falling. This turns the visual idea of a graph "going up" or "going down" into a precise, testable algebraic condition.
Increasing and decreasing — the definitions
A function $f$ is increasing on an interval $I$ if $x_1strictly increasing if $x_1decreasing and strictly decreasing. "Strictly" forbids any flat stretch; plain "increasing" (also called non-decreasing) allows the graph to stay level.
The monotonicity test
Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. On the open interval:
- If $f'(x)>0$ for all $x$, then $f$ is strictly increasing.
- If $f'(x)<0$ for all $x$, then $f$ is strictly decreasing.
- If $f'(x)=0$ for all $x$, then $f$ is constant.
The converse is one-directional: a function can be strictly increasing yet have $f'=0$ at an isolated point. For example $f(x)=x^3$ is strictly increasing on $\mathbb{R}$ even though $f'(0)=0$ — the derivative is zero at a single point but positive on either side, so the function never actually stops rising. This is why we test the sign on intervals, not the value at one point.
Method for finding intervals of increase and decrease
- Compute $f'(x)$ and factorise it as far as possible.
- Find the critical points where $f'(x)=0$ or $f'(x)$ is undefined; these split the domain into open subintervals.
- Test the sign of $f'$ in each subinterval (pick a convenient test point). A sign chart organises this neatly.
- Read off the answer: $f'>0$ marks increasing stretches, $f'<0$ decreasing ones. The points where $f'$ changes sign are the boundaries.
$$f'(x)>0 \ \text{on } I \ \Rightarrow\ f \text{ strictly increasing on } I \qquad f'(x)<0 \ \text{on } I \ \Rightarrow\ f \text{ strictly decreasing on } I$$
For functions built from $\sin$, $\cos$, $\log$ or $e^x$, the same recipe applies — you simply solve a trigonometric or transcendental inequality for the sign of $f'$. For trigonometric functions the answer is usually stated on a standard interval such as $[0,2\pi]$.
Deeper Insight — the derivative is a direction indicator: picture walking left to right along the graph. The first derivative is the slope under your feet: uphill ($f'>0$) means the output is rising, downhill ($f'<0$) means it is falling, and level ground ($f'=0$) is a momentary pause. Monotonicity questions are therefore nothing more than "where is the ground sloping up, and where down?" — and the only places the slope can switch sign are the critical points. That is exactly why the next topic, maxima and minima, is the same machinery: a peak is where uphill turns to downhill, a valley where downhill turns to uphill. Master the sign chart here and the extrema topic is already half done.
Find the intervals where $f(x)=x^2-4x+1$ is increasing/decreasing.
Solution$f'(x)=2x-4=0\Rightarrow x=2$. For $x<2,\ f'<0$ (decreasing); for $x>2,\ f'>0$ (increasing). So $f$ decreases on $(-\infty,2)$ and increases on $(2,\infty)$.
Answer: Decreasing on $(-\infty,2)$, increasing on $(2,\infty)$.
Show $f(x)=x^3+x$ is increasing on all of $\mathbb{R}$.
Solution$f'(x)=3x^2+1>0$ for every real $x$ (sum of a non-negative term and $1$). Hence $f$ is strictly increasing everywhere.
Answer: $f$ is strictly increasing on $\mathbb{R}$.
Find where $f(x)=x^3-3x$ is decreasing.
Solution$f'(x)=3x^2-3=3(x-1)(x+1)$. This is negative for $-1Answer: Decreasing on $(-1,1)$.
Is $f(x)=e^{x}$ increasing or decreasing?
Solution$f'(x)=e^{x}>0$ for all $x$, so $e^{x}$ is strictly increasing on $\mathbb{R}$.
Answer: Strictly increasing on $\mathbb{R}$.
Find the intervals in which $f(x)=2x^3-3x^2-36x+7$ is increasing and decreasing.
Solution$f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2)$. Critical points $x=-2,3$. Sign of $f'$: positive on $(-\infty,-2)$, negative on $(-2,3)$, positive on $(3,\infty)$.
Answer: Increasing on $(-\infty,-2)\cup(3,\infty)$; decreasing on $(-2,3)$.
Show that $f(x)=x^3-3x^2+3x-100$ is increasing on $\mathbb{R}$.
Solution$f'(x)=3x^2-6x+3=3(x^2-2x+1)=3(x-1)^2\ge 0$ for all $x$, and it is zero only at the single point $x=1$. Since $f'$ is positive on either side of $x=1$, $f$ never stops rising.
Answer: $f$ is increasing on all of $\mathbb{R}$.
Find the intervals of increase and decrease of $f(x)=x^4-2x^2$.
Solution$f'(x)=4x^3-4x=4x(x-1)(x+1)$. Critical points $x=-1,0,1$. Sign of $f'$: negative on $(-\infty,-1)$, positive on $(-1,0)$, negative on $(0,1)$, positive on $(1,\infty)$.
Answer: Increasing on $(-1,0)\cup(1,\infty)$; decreasing on $(-\infty,-1)\cup(0,1)$.
Prove that $f(x)=\sin x$ is strictly increasing on $\left(0,\dfrac{\pi}{2}\right)$ and strictly decreasing on $\left(\dfrac{\pi}{2},\pi\right)$.
Solution$f'(x)=\cos x$. On $\left(0,\dfrac{\pi}{2}\right)$, $\cos x>0$ so $f$ is strictly increasing; on $\left(\dfrac{\pi}{2},\pi\right)$, $\cos x<0$ so $f$ is strictly decreasing.
Answer: Increasing on $\left(0,\dfrac{\pi}{2}\right)$, decreasing on $\left(\dfrac{\pi}{2},\pi\right)$.
Find the intervals in which $f(x)=\dfrac{x}{x^2+1}$ is increasing or decreasing.
SolutionBy the quotient rule, $f'(x)=\dfrac{(x^2+1)(1)-x(2x)}{(x^2+1)^2}=\dfrac{1-x^2}{(x^2+1)^2}$. The denominator is always positive, so the sign of $f'$ follows $1-x^2=(1-x)(1+x)$: positive for $-1Answer: Increasing on $(-1,1)$; decreasing on $(-\infty,-1)\cup(1,\infty)$.
For what values of $x$ is $f(x)=\log(1+x)-\dfrac{x}{1+x}$ increasing, for $x>-1$?
Solution$f'(x)=\dfrac{1}{1+x}-\dfrac{(1+x)-x}{(1+x)^2}=\dfrac{1}{1+x}-\dfrac{1}{(1+x)^2}=\dfrac{(1+x)-1}{(1+x)^2}=\dfrac{x}{(1+x)^2}$. For $x>-1$ the denominator is positive, so $f'>0$ exactly when $x>0$.
Answer: Increasing on $(0,\infty)$ and decreasing on $(-1,0)$.
Find the interval in which $f(x)=x^2 e^{-x}$ is increasing.
Solution$f'(x)=2x e^{-x}+x^2(-e^{-x})=e^{-x}(2x-x^2)=x e^{-x}(2-x)$. Since $e^{-x}>0$ always, the sign of $f'$ follows $x(2-x)$, which is positive for $0Answer: Increasing on $(0,2)$; decreasing on $(-\infty,0)\cup(2,\infty)$.