Sign of the derivative and where a function increases or decreases, with practice problems

Q: Does the function always change direction where f'(x)=0?

Not necessarily. For y=x^3, f'(0)=0, but the function increases on both sides. Check whether the sign actually changes.

\begin{gathered} f'(x)>0 \Rightarrow \text{increasing}, \\[4pt] f'(x)<0 \Rightarrow \text{decreasing} \end{gathered}

The derivative is the slope at each point. The function increases where $f'(x)>0$ and decreases where $f'(x)<0$ .

A sign chart checks whether the sign changes at the points where $f'(x)=0$ .

f(x)=x³−3x. Orange where f'>0 (increasing), blue where f'<0 (decreasing). The boundaries are where f'=0.Drag the dot

Examples

01For which x is f'(x) positive? $\displaystyle f'(x)=3x-3$ AnswerHide

Answer $\displaystyle x>1$
Steps
1. $\displaystyle f'(x)$ $\displaystyle = 3x-3$ $\displaystyle = 3(x-1)$
2. Find where it is 0
3. $\displaystyle 3(x-1) = 0$ $\displaystyle \Rightarrow x = 1$
02If this is f'(x), on which interval is f(x) decreasing? $\displaystyle f'(x)=6x^2-24$ AnswerHide

Answer $\displaystyle -2<x<2$
Steps
1. $\displaystyle f'(x)$ $\displaystyle = 6x^2-24$ $\displaystyle = 6(x-2)(x+2)$
2. Find where it is 0
3. $\displaystyle 6(x-2)(x+2) = 0$ $\displaystyle \Rightarrow x = 2$
03If this is f'(x), on which interval is f(x) increasing? $\displaystyle f'(x)=4x-16$ AnswerHide

Answer $\displaystyle x>4$
Steps
1. $\displaystyle f'(x)$ $\displaystyle = 4x-16$ $\displaystyle = 4(x-4)$
2. Find where it is 0
3. $\displaystyle 4(x-4) = 0$ $\displaystyle \Rightarrow x = 4$
04f'(x) is positive up to the boundary □. What is □? $\displaystyle f'(x)=2x-10 \;\Rightarrow\; x>\square$ AnswerHide

Answer $\displaystyle 5$
Steps
1. $\displaystyle 2x-10$ $\displaystyle = 2(x-5)$
2. Find where it is 0
3. $\displaystyle 2(x-5) = 0$ $\displaystyle \Rightarrow x = 5$
05If this is f'(x), on which interval is f(x) decreasing? $\displaystyle f'(x)=3x^2-48$ AnswerHide

Answer $\displaystyle -4<x<4$
Steps
1. $\displaystyle f'(x)$ $\displaystyle = 3x^2-48$ $\displaystyle = 3(x-4)(x+4)$
2. Find where it is 0
3. $\displaystyle 3(x-4)(x+4) = 0$ $\displaystyle \Rightarrow x = 4$
06For which x is f'(x) positive? $\displaystyle f'(x)=4x-4$ AnswerHide

Answer $\displaystyle x>1$
Steps
1. $\displaystyle f'(x)$ $\displaystyle = 4x-4$ $\displaystyle = 4(x-1)$
2. Find where it is 0
3. $\displaystyle 4(x-1) = 0$ $\displaystyle \Rightarrow x = 1$

FAQ

Does the function always change direction where

f'(x)=0

Not necessarily. For

y=x^3

f'(0)=0

, but the function increases on both sides. Check whether the sign actually changes.

Published 2026-09-03 · Updated 2026-09-05