The first derivative tells us everything about where a function is rising or falling. If the derivative is positive, the function is increasing. If it's negative, the function is decreasing. This simple idea leads to the First Derivative Test for finding local maxima and minima.
🎯 In this section you will learn
- How to determine where a function is increasing or decreasing
- How to find critical points ($f'(x)=0$ or undefined)
- The First Derivative Test for local maxima and minima
- How to analyze the shape of a function using its derivative
📌 Increasing and Decreasing
$$ \text{If } f'(x) > 0 \text{ on an interval, then } f \text{ is } \textbf{increasing} \text{ on that interval} $$
$$ \text{If } f'(x) < 0 \text{ on an interval, then } f \text{ is } \textbf{decreasing} \text{ on that interval} $$
💡 Intuition
The derivative is the slope of the tangent line. Positive slope = function rising. Negative slope = function falling.
Find where $f(x) = x^3 - 3x$ is increasing and decreasing.
①
Find $f'(x)$
$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$
②
Find critical points
Set $f'(x)=0$: $x = 1$ and $x = -1$
③
Test intervals
On $(-\infty, -1)$: pick $x=-2$ → $f'(-2)=3(4-1)=9>0$ → increasing
On $(-1, 1)$: pick $x=0$ → $f'(0)=3(0-1)=-3<0$ → decreasing
On $(1, \infty)$: pick $x=2$ → $f'(2)=3(4-1)=9>0$ → increasing
📌 Critical Points
A critical point occurs where $f'(x) = 0$ or $f'(x)$ does not exist. These are the only places where a function can change from increasing to decreasing (or vice versa), and where local extrema can occur.
📌 First Derivative Test
Suppose $c$ is a critical point of $f$.
- If $f'$ changes from positive to negative at $c$, then $f$ has a local maximum at $c$.
- If $f'$ changes from negative to positive at $c$, then $f$ has a local minimum at $c$.
- If $f'$ does not change sign at $c$, then $f$ has no local extremum at $c$ (it's a saddle point).
Use the First Derivative Test to classify the critical points of $f(x) = x^3 - 3x$.
- From Example 1: $f'(x) = 3(x-1)(x+1)$
- At $x = -1$: $f'$ changes from positive to negative → local maximum at $x = -1$
- At $x = 1$: $f'$ changes from negative to positive → local minimum at $x = 1$
$f(-1) = -1 + 3 = 2$ (local max value). $f(1) = 1 - 3 = -2$ (local min value).
Analyze $f(x) = x^3$.
- $f'(x) = 3x^2$. Critical point at $x = 0$.
- For $x < 0$, $f'(x) > 0$; for $x > 0$, $f'(x) > 0$.
- The derivative does not change sign. There is no local extremum at $x = 0$ (it's an inflection point).
Find the critical points of $f(x) = \frac{x^2}{x-1}$.
①
Find $f'(x)$
Quotient Rule: $f'(x) = \frac{(2x)(x-1) - (x^2)(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$
②
Find where $f'(x)=0$ or undefined
$f'(x)=0$ when $x=0$ or $x=2$. $f'(x)$ undefined at $x=1$ (vertical asymptote).
③
Classify using sign analysis
Test intervals:
$(-\infty,0)$: pick $x=-1$ → $f'>0$ → increasing
$(0,1)$: pick $x=0.5$ → $f'<0$ → decreasing → local max at $x=0$
$(1,2)$: pick $x=1.5$ → $f'<0$ → decreasing
$(2,\infty)$: pick $x=3$ → $f'>0$ → increasing → local min at $x=2$
⚠️ Important Notes
- Critical points are candidates for local extrema, but not all critical points are extrema (e.g., $f(x)=x^3$ at $x=0$).
- Endpoints of a closed interval can also be extrema, but they are not critical points.
- When $f'(x)$ is undefined (e.g., cusp or vertical tangent), that point is still a critical point and should be tested.
🔍 Key Takeaways
- $f'(x) > 0$ ⇒ $f$ is increasing; $f'(x) < 0$ ⇒ $f$ is decreasing.
- Critical points: $f'(x) = 0$ or $f'(x)$ undefined.
- First Derivative Test: sign change of $f'$ determines local max/min.
- Always check the domain — vertical asymptotes and holes are not critical points (function not defined there).
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