Increasing and Decreasing Functions

Reading the sign of the derivative

Where $f'(x) > 0$, the function is increasing. Where $f'(x) < 0$, it's decreasing. Finding the intervals just means solving $f'(x) = 0$ for the boundary points, then checking the sign of $f'$ in each region between them.

Worked example

Find the intervals in which $f(x) = x^2 - 4x + 6$ is strictly increasing or decreasing.

Solution: $f'(x) = 2x - 4$, which is zero at $x=2$. For $x<2$, $f'(x)<0$; for $x>2$, $f'(x)>0$. So $f$ is strictly increasing on $(2, \infty)$ and strictly decreasing on $(-\infty, 2)$.

Worked example: a cubic with two turning points

Find the intervals in which $f(x) = 4x^3 - 6x^2 - 72x + 30$ is strictly increasing or decreasing.

Solution: $f'(x) = 12x^2 - 12x - 72 = 12(x-3)(x+2)$, zero at $x=-2$ and $x=3$. Testing a point in each of the three regions:

• At $x=-3$: $f'(-3) = 12(-6)(-1) = 72 > 0$
• At $x=0$: $f'(0) = 12(-3)(2) = -72 < 0$
• At $x=4$: $f'(4) = 12(1)(6) = 72 > 0$

So $f$ is strictly increasing on $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing on $(-2, 3)$.

The same test points laid out as a sign chart, the usual way to present this in an exam answer:

Common mistakes

• Testing only one region. $f'$ can change sign at every root, so check a point in every interval between consecutive roots.
• Forgetting points where $f'$ is undefined. Those split the number line too. For $f(x) = \dfrac{1}{x}$, $x = 0$ is a boundary even though $f'(x) = -\dfrac{1}{x^2}$ is never zero.
• Writing intervals with a union when they aren't connected. $f(x) = \dfrac{1}{x}$ is decreasing on $(-\infty, 0)$ and on $(0, \infty)$ separately, but not on their union: $f(-1) = -1 < f(1) = 1$.
• Mixing up strict and non-strict. $f(x) = x^3$ has $f'(0) = 0$ but is still strictly increasing everywhere. A single zero of $f'$ doesn't stop a function increasing.