The First Derivative Test

Using the derivative to classify critical points as maxima, minima, or neither

📊 The First Derivative Test

What is the First Derivative Test?

The First Derivative Test uses the sign of f(x)f'(x) to determine whether a critical point is a local maximum, local minimum, or neither.

💡 Key Idea: If the derivative changes from positive to negative at a critical point, you have a local max. If it changes from negative to positive, you have a local min!


Understanding the Signs of the Derivative

What f(x)f'(x) Tells Us

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

Think of it like driving a car:

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

The First Derivative Test (Formal Statement)

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

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

Case 1: Local Maximum

If ff' changes from positive to negative at cc:

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

Case 2: Local Minimum

If ff' changes from negative to positive at cc:

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

Case 3: Neither

If ff' does not change sign at cc:

  • Then f(c)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)=0f'(x) = 0 and find where f(x)f'(x) is undefined)

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

  • Mark critical points on a number line
  • Test the sign of f(x)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)=x33x29x+5f(x) = x^3 - 3x^2 - 9x + 5

Step 1: Find critical points

f(x)=3x26x9=3(x22x3)=3(x3)(x+1)f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)

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


Step 2: Create sign chart

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

Interval (,1)(-\infty, -1): Test x=2x = -2 f(2)=3(23)(2+1)=3(5)(1)=15>0f'(-2) = 3(-2-3)(-2+1) = 3(-5)(-1) = 15 > 0 ✓ POSITIVE

Interval (1,3)(-1, 3): Test x=0x = 0 f(0)=3(03)(0+1)=3(3)(1)=9<0f'(0) = 3(0-3)(0+1) = 3(-3)(1) = -9 < 0 ✓ NEGATIVE

Interval (3,)(3, \infty): Test x=4x = 4 f(4)=3(43)(4+1)=3(1)(5)=15>0f'(4) = 3(4-3)(4+1) = 3(1)(5) = 15 > 0 ✓ POSITIVE


Step 3: Sign chart

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

Step 4: Conclusions

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

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

Calculate the values:

  • f(1)=(1)33(1)29(1)+5=13+9+5=10f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10
  • f(3)=333(3)29(3)+5=272727+5=22f(3) = 3^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22

Answer:

  • Local maximum of 1010 at x=1x = -1
  • Local minimum of 22-22 at x=3x = 3

Creating Sign Charts

Method 1: Test Points

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

Tip: Use easy numbers like 0,1,1,10,100, 1, -1, 10, -10

Method 2: Factored Form

If f(x)=a(xr1)(xr2)f'(x) = a(x - r_1)(x - r_2)\cdots, analyze each factor's sign.

Example: f(x)=(x+2)(x1)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 ff is increasing or decreasing:

From the Sign Chart

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

Example from above:

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

Special Cases

Case 1: Inflection Point with Horizontal Tangent

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

Example: f(x)=x3f(x) = x^3 at x=0x = 0

  • f(x)=3x2f'(x) = 3x^2, so f(0)=0f'(0) = 0
  • f(x)>0f'(x) > 0 for all x0x \neq 0 (no sign change!)
  • x=0x = 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)=xf(x) = |x| at x=0x = 0

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

⚠️ Common Mistakes

Mistake 1: Testing AT the Critical Point

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

Mistake 2: Wrong Sign Interpretation

The sign of ff', not ff, 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)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)=0f''(c) = 0 ❌ Doesn't give increasing/decreasing info


Quick Reference Chart

| Sign Change | Result | |-------------|--------| | + → − | Local MAX ⬆️⬇️ | | − → + | Local MIN ⬇️⬆️ | | + → + | Neither (increasing through) | | − → − | Neither (decreasing through) |


📝 Practice Strategy

  1. Find and factor f(x)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 ff values at extrema for complete answer

📚 Practice Problems

1Problem 1medium

Question:

Use the First Derivative Test to find all local extrema of f(x)=x44x3f(x) = x^4 - 4x^3.

💡 Show Solution

Step 1: Find the derivative

f(x)=x44x3f(x) = x^4 - 4x^3

f(x)=4x312x2=4x2(x3)f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)


Step 2: Find critical points

4x2(x3)=04x^2(x - 3) = 0

x=0x = 0 or x=3x = 3


Step 3: Create sign chart

Test intervals: (,0)(-\infty, 0), (0,3)(0, 3), (3,)(3, \infty)

Test x=1x = -1 (in (,0)(-\infty, 0)): f(1)=4(1)2(13)=4(1)(4)=16<0f'(-1) = 4(-1)^2(-1-3) = 4(1)(-4) = -16 < 0 ✓ NEGATIVE

Test x=1x = 1 (in (0,3)(0, 3)): f(1)=4(1)2(13)=4(1)(2)=8<0f'(1) = 4(1)^2(1-3) = 4(1)(-2) = -8 < 0 ✓ NEGATIVE

Test x=4x = 4 (in (3,)(3, \infty)): f(4)=4(4)2(43)=4(16)(1)=64>0f'(4) = 4(4)^2(4-3) = 4(16)(1) = 64 > 0 ✓ POSITIVE


Step 4: Analyze sign chart

        0         3
   ----  |  ----  |  ++++
                  ▲

At x=0x = 0: − to − (no change) → NEITHER (inflection point)

At x=3x = 3: − to + → LOCAL MINIMUM


Step 5: Calculate the minimum value

f(3)=344(3)3=81108=27f(3) = 3^4 - 4(3)^3 = 81 - 108 = -27

Answer: Local minimum of 27-27 at x=3x = 3. No local maximum. (x=0x = 0 is an inflection point with horizontal tangent.)

2Problem 2hard

Question:

Find the intervals where g(x)=x+4xg(x) = x + \frac{4}{x} is increasing and decreasing. Identify all local extrema.

💡 Show Solution

Step 1: Find the derivative

g(x)=x+4x1g(x) = x + 4x^{-1}

g(x)=14x2=14x2g'(x) = 1 - 4x^{-2} = 1 - \frac{4}{x^2}


Step 2: Find critical points

Set g(x)=0g'(x) = 0:

14x2=01 - \frac{4}{x^2} = 0

4x2=1\frac{4}{x^2} = 1

x2=4x^2 = 4

x=±2x = \pm 2

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

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


Step 3: Create sign chart

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

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

Test x=3x = -3: g(3)=149=59>0g'(-3) = 1 - \frac{4}{9} = \frac{5}{9} > 0 ✓ POSITIVE

Test x=1x = -1: g(1)=141=3<0g'(-1) = 1 - \frac{4}{1} = -3 < 0 ✓ NEGATIVE

Test x=1x = 1: g(1)=141=3<0g'(1) = 1 - \frac{4}{1} = -3 < 0 ✓ NEGATIVE

Test x=3x = 3: g(3)=149=59>0g'(3) = 1 - \frac{4}{9} = \frac{5}{9} > 0 ✓ POSITIVE


Step 4: Sign chart

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

Step 5: Classify critical points

At x=2x = -2: + to − → LOCAL MAXIMUM

At x=2x = 2: − to + → LOCAL MINIMUM

Calculate values:

  • g(2)=2+42=22=4g(-2) = -2 + \frac{4}{-2} = -2 - 2 = -4
  • g(2)=2+42=2+2=4g(2) = 2 + \frac{4}{2} = 2 + 2 = 4

Answer:

  • Increasing on (,2)(-\infty, -2) and (2,)(2, \infty)
  • Decreasing on (2,0)(-2, 0) and (0,2)(0, 2)
  • Local maximum of 4-4 at x=2x = -2
  • Local minimum of 44 at x=2x = 2

3Problem 3expert

Question:

Use the First Derivative Test on h(x)=x2/3(x4)h(x) = x^{2/3}(x-4) to find and classify all critical points.

💡 Show Solution

Step 1: Expand and find derivative

h(x)=x2/3(x4)=x5/34x2/3h(x) = x^{2/3}(x-4) = x^{5/3} - 4x^{2/3}

h(x)=53x2/3423x1/3h'(x) = \frac{5}{3}x^{2/3} - 4 \cdot \frac{2}{3}x^{-1/3}

h(x)=53x2/383x1/3h'(x) = \frac{5}{3}x^{2/3} - \frac{8}{3}x^{-1/3}


Step 2: Combine using common denominator

h(x)=5x2/3x1/383x1/3=5x83x1/3h'(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)=0h'(x) = 0: 5x8=05x - 8 = 0x=85x = \frac{8}{5}

h(x)h'(x) undefined: x1/3=0x^{1/3} = 0x=0x = 0

Critical points: x=0x = 0 and x=85x = \frac{8}{5}


Step 4: Test signs

Test x=1x = -1 (in (,0)(-\infty, 0)): h(1)=5(1)83(1)1/3=133(1)=133>0h'(-1) = \frac{5(-1)-8}{3(-1)^{1/3}} = \frac{-13}{3(-1)} = \frac{-13}{-3} > 0 ✓ POSITIVE

Test x=1x = 1 (in (0,85)(0, \frac{8}{5})): h(1)=5(1)83(1)1/3=33=1<0h'(1) = \frac{5(1)-8}{3(1)^{1/3}} = \frac{-3}{3} = -1 < 0 ✓ NEGATIVE

Test x=2x = 2 (in (85,)(\frac{8}{5}, \infty)): h(2)=5(2)83(2)1/3=2323>0h'(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=0x = 0: + to − → LOCAL MAXIMUM

At x=85x = \frac{8}{5}: − to + → LOCAL MINIMUM


Step 6: Calculate values

h(0)=02/3(04)=0h(0) = 0^{2/3}(0-4) = 0

h(85)=(85)2/3(854)h\left(\frac{8}{5}\right) = \left(\frac{8}{5}\right)^{2/3}\left(\frac{8}{5}-4\right)

=(85)2/3(125)= \left(\frac{8}{5}\right)^{2/3}\left(-\frac{12}{5}\right)

=125452/33.07= -\frac{12}{5} \cdot \frac{4}{5^{2/3}} \approx -3.07

Answer:

  • Local maximum of 00 at x=0x = 0
  • Local minimum of 125(85)2/33.07-\frac{12}{5}\left(\frac{8}{5}\right)^{2/3} \approx -3.07 at x=85x = \frac{8}{5}