Increasing and decreasing functions

Formal definitions. A function $f$ is increasing on an interval $I$ if for any two points $x_1<x_2$ in $I$, $f(x_1)\le f(x_2)$ — strictly increasing if the inequality is strict, $f(x_1)<f(x_2)$. It is decreasing if $f(x_1)\ge f(x_2)$ for $x_1<x_2$, strictly decreasing if $f(x_1)>f(x_2)$. These definitions talk only about the values of $f$ itself, with no mention of a derivative — the derivative is the tool used to check them quickly.

The derivative test. For a differentiable function $f$ on an interval $I$:

where: $f'(x)$ is the instantaneous rate of change of $f$ at $x$ — geometrically, the slope of the tangent line at that point. A positive slope means the tangent points "uphill," so the curve is climbing; a negative slope means the tangent points "downhill."

The working method. First, differentiate $f$ to get $f'(x)$. Next, solve $f'(x)=0$ to locate the critical points, which split the domain into subintervals. Then pick one convenient test value inside each subinterval and evaluate the sign of $f'(x)$ there — the sign is guaranteed to stay constant across the whole subinterval, because a continuous derivative cannot change sign without passing through zero somewhere in between. Finally, read off the conclusion: subintervals where $f'(x)>0$ are increasing, and where $f'(x)<0$ are decreasing. Exactly the same sign-chart habit reappears when concavity is studied using the second derivative, so it is worth building cleanly here.

Bounding a derivative without factorising. Not every function has a derivative that factors into neat linear pieces. When $f'(x)$ involves a bounded expression such as $\sin x$ or $\cos x$, the sign can often be pinned down directly from the known range of that expression, without solving $f'(x)=0$ at all. For instance, $\cos x$ always lies between $-1$ and $1$, so an expression like $1-\cos x$ can never dip below zero — it is squeezed to be non-negative for every real $x$, and equals zero only where $\cos x$ touches its maximum value of $1$. This bounding technique is a common alternative to the critical-point method whenever trigonometric or other bounded terms appear in $f'(x)$, and it proves monotonicity over the whole real line in one line of reasoning rather than a sign chart with infinitely many pieces.

Find the intervals in which $f(x) = 2x^3-15x^2+36x+1$ is increasing and decreasing, and confirm the classification at $x=2.5$ using the function values themselves.

Step 1. Differentiate: $f'(x) = 6x^2-30x+36 = 6(x^2-5x+6)$.

Step 2. Factorise and locate the critical points: $f'(x) = 6(x-2)(x-3)$, so $f'(x)=0$ at $x=2$ and $x=3$. These split the real line into three test intervals: $(-\infty,2)$, $(2,3)$, $(3,\infty)$.

Step 3. Test the sign of $f'(x)$ in each interval using a convenient value. On $(-\infty,2)$, try $x=0$: $f'(0)=6(-2)(-3)=36>0$. On $(2,3)$, try $x=2.5$: $f'(2.5)=6(0.5)(-0.5)=-1.5<0$. On $(3,\infty)$, try $x=4$: $f'(4)=6(2)(1)=12>0$.

Step 4. Conclude and sanity-check with actual values. $f$ is increasing on $(-\infty,2)\cup(3,\infty)$ and decreasing on $(2,3)$. To confirm the decreasing middle interval, compute $f(2)=2(8)-15(4)+36(2)+1=16-60+72+1=29$ and $f(3)=2(27)-15(9)+36(3)+1=54-135+108+1=28$. Since $f(3)=28<29=f(2)$, the function value genuinely dropped across $(2,3)$, matching the sign test.

f is increasing on $(-\infty,2)\cup(3,\infty)$ and decreasing on $(2,3)$, confirmed by $f(2)=29 > f(3)=28$.

Competitive stretch

Show that $g(x)=x-\sin x$ is increasing for every real $x$, without solving $g'(x)=0$ by factorising.

Step 1. Differentiate: $g'(x)=1-\cos x$.

Step 2. Bound the sign using the range of cosine, exactly as described above. Since $-1\le\cos x\le1$ for every real $x$, we get $1-\cos x\ge 1-1=0$ for all $x$, with equality only at the isolated points where $\cos x=1$, that is, $x=2n\pi$ for integer $n$.

Step 3. Interpret the isolated zeros. Because $g'(x)\ge0$ everywhere and equals zero only at scattered points rather than across any interval, the function never has a genuine flat stretch, so $g$ is strictly increasing on all of $\mathbb{R}$.

• Forgetting to test the sign of $f'(x)$ separately in each subinterval, instead assuming the signs simply alternate — unreliable with repeated or higher-order factors.
• Treating $f'(x)=0$ as automatically "neither increasing nor decreasing" without checking whether the zero is isolated (the function can still be strictly increasing overall) or marks a genuine flat or turning region.
• Sign errors while factorising a quadratic derivative, which swap the increasing and decreasing intervals in the final answer.
• Choosing a test value that accidentally lands on another critical point or an undefined point of the function, giving a misleading sign reading for that whole subinterval.

Ask the tutor: Not sure why an isolated zero of $f'(x)$ doesn't break strict monotonicity, or want a walkthrough of picking safe test values with a graph alongside — ask the AI tutor.