• $f'(x) > 0$: Function is increasing (going up)
• $f'(x) < 0$: Function is decreasing (going down)
• $f'(x) = 0$: Function has a horizontal tangent

Think of it like driving a car:

• $f'(x) > 0$: driving uphill ⬆️
• $f'(x) < 0$: driving downhill ⬇️
• $f'(x) = 0$: at the top or bottom of a hill

The First Derivative Test (Formal Statement)

Let $c$ be a critical point of $f$ (so $f'(c) = 0$ or $f'(c)$ undefined).

Test the sign of $f'(x)$ on intervals around $c$:

Case 1: Local Maximum

If $f'$ changes from positive to negative at $c$:

• $f'(x) > 0$ for $x < c$ (increasing before)
• $f'(x) < 0$ for $x > c$ (decreasing after)
• Then $f(c)$ is a local maximum 📈➡️📉

Case 2: Local Minimum

If $f'$ changes from negative to positive at $c$:

• $f'(x) < 0$ for $x < c$ (decreasing before)
• $f'(x) > 0$ for $x > c$ (increasing after)
• Then $f(c)$ is a local minimum 📉➡️📈

Case 3: Neither

If $f'$ does not change sign at $c$:

• Then $f(c)$ is neither a max nor a min
• Could be an inflection point

Step-by-Step Process

How to Apply the First Derivative Test

Step 1: Find all critical points (solve $f'(x) = 0$ and find where $f'(x)$ is undefined)

Step 2: Create a sign chart for $f'(x)$

• Mark critical points on a number line
• Test the sign of $f'(x)$ in each interval

Step 3: Analyze the sign changes

• Positive → Negative = Local MAX
• Negative → Positive = Local MIN
• No change = Neither

Step 4: State your conclusions

Example 1: Complete Analysis

Classify the critical points of $f(x) = x^3 - 3x^2 - 9x + 5$

Step 1: Find critical points

$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$

Critical points: $x = -1$ and $x = 3$

Step 2: Create sign chart

Test intervals: $(-\infty, -1)$, $(-1, 3)$, $(3, \infty)$

Interval $(-\infty, -1)$: Test $x = -2$ $f'(-2) = 3(-2-3)(-2+1) = 3(-5)(-1) = 15 > 0$ ✓ POSITIVE

Interval $(-1, 3)$: Test $x = 0$ $f'(0) = 3(0-3)(0+1) = 3(-3)(1) = -9 < 0$ ✓ NEGATIVE

Interval $(3, \infty)$: Test $x = 4$ $f'(4) = 3(4-3)(4+1) = 3(1)(5) = 15 > 0$ ✓ POSITIVE

Step 3: Sign chart

        -1        3
   ++++  |  ----  |  ++++
         ▼        ▲

Step 4: Conclusions

At $x = -1$: $f'$ goes from + to − → LOCAL MAXIMUM

At $x = 3$: $f'$ goes from − to + → LOCAL MINIMUM

Calculate the values:

• $f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$
• $f(3) = 3^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22$

Answer:

• Local maximum of $10$ at $x = -1$
• Local minimum of $-22$ at $x = 3$

Creating Sign Charts

Method 1: Test Points

Pick any value in each interval and substitute into $f'(x)$.

Tip: Use easy numbers like $0, 1, -1, 10, -10$

Method 2: Factored Form

If $f'(x) = a(x - r_1)(x - r_2)\cdots$, analyze each factor's sign.

Example: $f'(x) = (x+2)(x-1)$

Interval      (x+2)  (x-1)  Product
(-∞, -2)        −      −       +
(-2, 1)         +      −       −
(1, ∞)          +      +       +

Increasing and Decreasing Intervals

The First Derivative Test also tells us where $f$ is increasing or decreasing:

From the Sign Chart

• Where $f'(x) > 0$: $f$ is increasing
• Where $f'(x) < 0$: $f$ is decreasing

Example from above:

• $f$ is increasing on $(-\infty, -1)$ and $(3, \infty)$
• $f$ is decreasing on $(-1, 3)$

Special Cases

Case 1: Inflection Point with Horizontal Tangent

If $f'(c) = 0$ but the sign doesn't change, it's usually an inflection point.

Example: $f(x) = x^3$ at $x = 0$

• $f'(x) = 3x^2$, so $f'(0) = 0$
• $f'(x) > 0$ for all $x \neq 0$ (no sign change!)
• $x = 0$ is an inflection point, not an extremum

Case 2: Multiple Critical Points in a Row

Handle each one separately using the test.

Case 3: Derivative Undefined

The test still works! Just check the sign change around the point.

Example: $f(x) = |x|$ at $x = 0$

• $f'(x) < 0$ for $x < 0$
• $f'(x) > 0$ for $x > 0$
• Changes from − to + → local minimum at $x = 0$ ✓

⚠️ Common Mistakes

Mistake 1: Testing AT the Critical Point

❌ Don't test $f'(c)$ where $c$ is the critical point ✅ Test points on either SIDE of $c$

Mistake 2: Wrong Sign Interpretation

The sign of $f'$, not $f$, determines increasing/decreasing!

Mistake 3: Forgetting to Check All Critical Points

Each critical point needs its own analysis.

Mistake 4: Not Using Factored Form

Factor $f'(x)$ completely to make sign analysis easier!

First Derivative Test vs. Other Methods

When to Use the First Derivative Test

✅ Always works (if derivative exists on either side) ✅ Good for understanding overall behavior ✅ Tells you increasing/decreasing intervals ✅ Works when second derivative is hard to compute

When to Use the Second Derivative Test (coming next!)

✅ Faster for simple cases ✅ Only need to evaluate at one point ❌ Fails if $f''(c) = 0$ ❌ Doesn't give increasing/decreasing info

Quick Reference Chart

📝 Practice Strategy

1. Find and factor $f'(x)$ completely
2. Mark critical points on a number line
3. Test one point in each interval (use easy numbers)
4. Record signs carefully: + for positive, − for negative
5. Look for sign changes: + to − is max, − to + is min
6. State intervals: Where increasing? Where decreasing?
7. Calculate $f$ values at extrema for complete answer

Step 2: Find critical points

Set $g'(x) = 0$:

$1 - \frac{4}{x^2} = 0$

$\frac{4}{x^2} = 1$

$x^2 = 4$

$x = \pm 2$

Also, $g'(x)$ is undefined at $x = 0$, but $g(0)$ is also undefined, so $x = 0$ is NOT a critical point (not in domain).

Critical points: $x = -2$ and $x = 2$

Step 3: Create sign chart

Test intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 2)$, $(2, \infty)$

Note: $x = 0$ is not in the domain but divides the intervals.

Test $x = -3$: $g'(-3) = 1 - \frac{4}{9} = \frac{5}{9} > 0$ ✓ POSITIVE

Test $x = -1$: $g'(-1) = 1 - \frac{4}{1} = -3 < 0$ ✓ NEGATIVE

Test $x = 1$: $g'(1) = 1 - \frac{4}{1} = -3 < 0$ ✓ NEGATIVE

Test $x = 3$: $g'(3) = 1 - \frac{4}{9} = \frac{5}{9} > 0$ ✓ POSITIVE

Step 4: Sign chart

       -2    0    2
   ++++  | -- | --  | ++++
         ▼      ▲

Step 5: Classify critical points

At $x = -2$: + to − → LOCAL MAXIMUM

At $x = 2$: − to + → LOCAL MINIMUM

Calculate values:

• $g(-2) = -2 + \frac{4}{-2} = -2 - 2 = -4$
• $g(2) = 2 + \frac{4}{2} = 2 + 2 = 4$

Answer:

• Increasing on $(-\infty, -2)$ and $(2, \infty)$
• Decreasing on $(-2, 0)$ and $(0, 2)$
• Local maximum of $-4$ at $x = -2$
• Local minimum of $4$ at $x = 2$

$h'(x) = \frac{5}{3}x^{2/3} - 4 \cdot \frac{2}{3}x^{-1/3}$

$h'(x) = \frac{5}{3}x^{2/3} - \frac{8}{3}x^{-1/3}$

Step 2: Combine using common denominator

$h'(x) = \frac{5x^{2/3} \cdot x^{1/3} - 8}{3x^{1/3}} = \frac{5x - 8}{3x^{1/3}}$

Step 3: Find critical points

$h'(x) = 0$: $5x - 8 = 0$ → $x = \frac{8}{5}$

$h'(x)$ undefined: $x^{1/3} = 0$ → $x = 0$

Critical points: $x = 0$ and $x = \frac{8}{5}$

Step 4: Test signs

Test $x = -1$ (in $(-\infty, 0)$): $h'(-1) = \frac{5(-1)-8}{3(-1)^{1/3}} = \frac{-13}{3(-1)} = \frac{-13}{-3} > 0$ ✓ POSITIVE

Test $x = 1$ (in $(0, \frac{8}{5})$): $h'(1) = \frac{5(1)-8}{3(1)^{1/3}} = \frac{-3}{3} = -1 < 0$ ✓ NEGATIVE

Test $x = 2$ (in $(\frac{8}{5}, \infty)$): $h'(2) = \frac{5(2)-8}{3(2)^{1/3}} = \frac{2}{3\sqrt[3]{2}} > 0$ ✓ POSITIVE

Step 5: Sign chart

        0      8/5
   ++++  |  ----  |  ++++
         ▼        ▲

At $x = 0$: + to − → LOCAL MAXIMUM

At $x = \frac{8}{5}$: − to + → LOCAL MINIMUM

Step 6: Calculate values

$h(0) = 0^{2/3}(0-4) = 0$

$h\left(\frac{8}{5}\right) = \left(\frac{8}{5}\right)^{2/3}\left(\frac{8}{5}-4\right)$

$= \left(\frac{8}{5}\right)^{2/3}\left(-\frac{12}{5}\right)$

$= -\frac{12}{5} \cdot \frac{4}{5^{2/3}} \approx -3.07$

Answer:

• Local maximum of $0$ at $x = 0$
• Local minimum of $-\frac{12}{5}\left(\frac{8}{5}\right)^{2/3} \approx -3.07$ at $x = \frac{8}{5}$