Learn how to use derivatives to analyze function behavior, find turning points, and determine where functions increase or decrease.
Understanding when a function increases or decreases is fundamental in calculus and has numerous real-world applications. Think about:
A function is increasing on an interval if, as you move from left to right (increasing $x$), the $y$-values also increase.
Mathematically:
For $x_1 < x_2$, if $f(x_1) < f(x_2)$, then $f$ is increasing
A function is decreasing on an interval if, as you move from left to right (increasing $x$), the $y$-values decrease.
Mathematically:
For $x_1 < x_2$, if $f(x_1) > f(x_2)$, then $f$ is decreasing
Real-World Examples:
The derivative $f'(x)$ tells us the gradient (slope) of the tangent line to the function at any point. This gradient reveals whether the function is increasing or decreasing!
When the derivative is positive, the slope is positive (going uphill), so the function is increasing.
When the derivative is negative, the slope is negative (going downhill), so the function is decreasing.
When the derivative is zero, the slope is horizontal. This is a turning point (also called critical point or stationary point).
Remember that the derivative $f'(x)$ represents the rate of change of the function. It's like the speedometer in a car:
Critical points, also called turning points or stationary points, are where the function "changes direction" — from increasing to decreasing, or vice versa.
Use the differentiation rules to find $f'(x)$
Solve the equation $f'(x) = 0$ to find the $x$-coordinates of critical points
Substitute the $x$-values back into $f(x)$ to find the complete coordinates
Important Note:
Finding where $f'(x) = 0$ gives you potential turning points. To confirm they are actually maxima or minima (and not points of inflection), you need to test the intervals around them!
Step 1: Find $f'(x)$
Step 2: Set $f'(x) = 0$ and solve
Result:
The critical points occur at $x = -1$ and $x = 1$. These are where the function has turning points!
Once you've found the critical points, you need to determine whether the function is increasing or decreasing between and around these points. We do this by testing the sign of the derivative in each interval.
These are the $x$-values where $f'(x) = 0$. They divide the number line into intervals.
Mark the critical points on a number line to create intervals
Pick any value within each interval (doesn't have to be special, just any number in that range)
Check if the derivative is positive or negative
If $f'(x) > 0$: function is increasing
If $f'(x) < 0$: function is decreasing
We have $f'(x) = 3x^2 - 3$ and critical points at $x = -1$ and $x = 1$
Number Line with Intervals:
(−∞, −1) (−1, 1) (1, ∞)
────────●────────●────────→
-1 1Test each interval:
Interval: $x < -1$
Test point: $x = -2$
✓ Increasing
Interval: $-1 < x < 1$
Test point: $x = 0$
✗ Decreasing
Interval: $x > 1$
Test point: $x = 2$
✓ Increasing
Conclusion:
💡 Pro Tip: Factored Form
If you can factor the derivative, you can often determine the sign without calculating specific values. For example, if $f'(x) = 3(x + 1)(x - 1)$, you can use a sign chart based on the factors!
Not all turning points are the same! A turning point can be a maximum (peak), a minimum (valley), or sometimes neither (point of inflection). Here's how to tell the difference.
Sign pattern:
$f'(x)$: + → 0 → −
The derivative changes from positive to negative
Function was increasing, reached a peak, then started decreasing
Sign pattern:
$f'(x)$: − → 0 → +
The derivative changes from negative to positive
Function was decreasing, reached a valley, then started increasing
Sign pattern:
$f'(x)$: + → 0 → + or − → 0 → −
The derivative does not change sign (stays positive or stays negative)
Not covered in detail for Grade 12 CAPS, but good to know it exists!
From our interval testing, we found:
Increasing Decreasing Increasing
(+) (−) (+)
────────●────────●────────→
-1 1
MAX MINAt $x = -1$:
$f'(x)$ changes from + to −
∴ Maximum turning point
At $x = 1$:
$f'(x)$ changes from − to +
∴ Minimum turning point
Important Terminology:
Test $x = -1$
$g'(-1) = 6(-1)^2 - 12(-1)$
$= 6 + 12 = 18 > 0$
Increasing
Test $x = 1$
$g'(1) = 6(1)^2 - 12(1)$
$= 6 - 12 = -6 < 0$
Decreasing
Test $x = 3$
$g'(3) = 6(3)^2 - 12(3)$
$= 54 - 36 = 18 > 0$
Increasing
At $x = 0$:
Sign changes: + → −
Maximum
At $x = 2$:
Sign changes: − → +
Minimum
Final Answer:
Find derivative:
$$h'(x) = -3x^2 + 12$$Set $h'(x) = 0$:
$$\begin{aligned} -3x^2 + 12 &= 0 \\ -3x^2 &= -12 \\ x^2 &= 4 \\ x &= \pm 2 \end{aligned}$$Test intervals:
Test $x = -3$:
$h'(-3) = -3(9) + 12 = -15$
Decreasing
Test $x = 0$:
$h'(0) = 0 + 12 = 12$
Increasing
Test $x = 3$:
$h'(3) = -3(9) + 12 = -15$
Decreasing
Answer:
Explore how derivatives relate to function behavior with this interactive tool.
Select Function:
Observations:
Test your understanding with these practice problems.
Test your understanding with these multiple-choice questions.
Continue your learning journey with more interactive calculus lessons aligned with the CAPS curriculum.