Increasing and decreasing functions

What you will be able to do

• define increasing, decreasing and strictly monotonic functions;
• use the sign of $f'$ to find where a function rises and falls;
• find the intervals of increase and decrease of a polynomial or other function;
• use monotonic behaviour to count roots and to prove an inequality.

Rising and falling

Walk along a graph from left to right. If you climb, the function is increasing. If you descend, it is decreasing. In symbols, for $x_1 < x_2$ in an interval $I$:

• $f$ is increasing on $I$ if $f(x_1) \le f(x_2)$;
• $f$ is strictly increasing if $f(x_1) < f(x_2)$;
• $f$ is decreasing on $I$ if $f(x_1) \ge f(x_2)$, and strictly decreasing if $f(x_1) > f(x_2)$.

A word of care. NCERT uses these definitions, so a constant function counts as increasing. A-Level, IB and AP usually say increasing when they mean strictly increasing. In this course we write strictly whenever we mean it, so there is no doubt. A function that is increasing everywhere, or decreasing everywhere, is called monotonic.

The sign of the derivative

The derivative is the gradient of the graph. A positive gradient means the graph climbs. This gives the test.

Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$.

• If $f'(x) > 0$ for all $x$ in $(a, b)$, then $f$ is strictly increasing on $[a, b]$.
• If $f'(x) < 0$ for all $x$ in $(a, b)$, then $f$ is strictly decreasing on $[a, b]$.

The graph above has $f'(x) = 3(x - 1)(x - 3)$. It is positive before $1$, negative between $1$ and $3$, and positive after $3$.

Finding the intervals

1. Differentiate. Find $f'(x)$.
2. Solve $f'(x) = 0$. These points cut the number line into gaps. Also note any point where $f'$ does not exist.
3. Test the sign of $f'$ in each gap. Pick one test value in each gap.
4. Write the answer. Include the end points where $f$ is continuous.

For $f(x) = x^3 - 6x^2 + 9x + 2$, we have $f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$.

So $f$ is increasing on $(-\infty, 1]$ and $[3, \infty)$, and decreasing on $[1, 3]$.

Strictly monotonic functions

Take $f(x) = x^3$. Its derivative $3x^2$ is $0$ at $x = 0$. Does the graph stop rising there? No. It flattens for an instant and keeps climbing. The rule is:

The rule. If $f'(x) \ge 0$ on an interval, and $f'(x) = 0$ only at isolated points, then $f$ is strictly increasing on the interval.

So $x^3$ is strictly increasing on all of $\mathbb{R}$. Another example is $f(x) = x + \sin x$. Its derivative $1 + \cos x$ is $0$ only at isolated points $x = \pi, 3\pi, \dots$, so $f$ is strictly increasing even though its graph has flat moments.

A strictly monotonic function never takes the same value twice. Two things follow.

• An equation $f(x) = k$ has at most one root on that interval.
• The function is one-one, so it has an inverse.

Proving an inequality

To show that $x > \sin x$ for $x > 0$, let $g(x) = x - \sin x$. Then $g'(x) = 1 - \cos x \ge 0$, with zeros only at isolated points. So $g$ is strictly increasing. Since $g(0) = 0$, we get $g(x) > 0$ for $x > 0$.

A worked example

Find the intervals on which $f(x) = x^2 e^{-x}$ increases and decreases.

1. Differentiate. $f'(x) = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2 - x)$.
2. Solve $f'(x) = 0$. The factor $e^{-x}$ is never $0$, so $x = 0$ or $x = 2$.
3. Test the gaps. For $x < 0$: $x$ is negative and $2 - x$ is positive, so $f' < 0$. For $0 < x < 2$: both are positive, so $f' > 0$. For $x > 2$: $2 - x$ is negative, so $f' < 0$.
4. Answer. Decreasing on $(-\infty, 0]$ and $[2, \infty)$, increasing on $[0, 2]$.

Practise with new numbers

Try it in Python

The program differentiates a polynomial, finds where the derivative is $0$ and tests a point in each gap. It reports whether the function is increasing or decreasing there.

Common mistakes

• Testing the sign of $f$ instead of $f'$. The graph rises where $f'$ is positive, whatever the value of $f$.
• Thinking $f'(c) = 0$ means $f$ stops rising. For $x^3$ it keeps rising through $x = 0$.
• Joining two intervals that are separated. $f$ is increasing on $(-\infty, 1]$ and on $[3, \infty)$. It is not increasing on their union.
• Forgetting a point where $f'$ does not exist. It can also split the number line.
• Writing a wrong direction at an end point. The test value, not the end point, tells the direction.

Remember these

• $f' > 0$ on an interval means strictly increasing. $f' < 0$ means strictly decreasing.
• Solve $f'(x) = 0$, test each gap, and write the intervals.
• $f' \ge 0$ with isolated zeros still gives a strictly increasing function.
• A strictly monotonic function has at most one root of $f(x) = k$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Is $f(x) = 7 - 3x$ increasing or decreasing? Is it strict?

2. Find the interval on which $f(x) = x^2 - 6x + 5$ is decreasing.

3. Show that $f(x) = x^3 + 3x$ is strictly increasing on $\mathbb{R}$.

4. Why is $f(x) = x^3$ strictly increasing although $f'(0) = 0$?

5. Show that $\cos x$ is strictly decreasing on $(0, \pi)$.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives), the section on increasing and decreasing functions.
• OpenStax, Calculus Volume 1, Section 4.5 (Derivatives and the Shape of a Graph).