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Test for a Minimum

Mathematics · FE Reference Handbook section

Mathematics
2 formulas
10 exam-style examples
~49 min
All Mathematics lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Maximum of a cost function

Total cost of a haul road is C(x) = x³ − 9x² + 24x + 5 ($1,000) with x the haul distance in km. Locate the local maximum and its value.

Given

  • C(x)=x3−9x2+24x+5C(x) = x^{3} - 9x^{2} + 24x + 5

Find

x at the local maximum and C at that point

Start with the thinking

  • Critical points come from C′(x) = 0; the second derivative classifies them.
  • A local maximum needs C″ < 0.
maxminxyC(x) with both critical points

Figure 1 — schematic for Maximum of a cost function

Step-by-step solution

  1. First derivative

    C′(x)=3x2−18x+24C'(x) = 3x^{2} - 18x + 24
  2. Set to zero

    3x2−18x+24=0,orx2−6x+8=03x^{2} - 18x + 24 = 0, or x^{2} - 6x + 8 = 0
  3. Factor

    (x−2)(x−4)=0,sox=2andx=4(x - 2)(x - 4) = 0, so x = 2 and x = 4
  4. Second derivative

    C″(x)=6x−18C″(x) = 6x - 18
  5. Classify

    C″(2)=−6<0(maximum);C″(4)=+6>0(minimum)C″(2) = -6 < 0 (maximum); C″(4) = +6 > 0 (minimum)
  6. Value — C(2) = 8 − 36 + 48 + 5 = 25.0 ($1,000)

Answer:

Local maximum at x = 2 km, C = $25,000

Why the other options are there

  • x = 4 km (minimum reported)
  • C = 21 (arithmetic slip on 8 − 36 + 48)

Reference: FE Reference Handbook — Mathematics — Differential calculus

Example 2
Derivative evaluated at a point — Test for a Minimum

For f(x) = 5x³ − 7x² + 7x − 5, what is f ′(1)?

Given

  • f(x)=5x3−7x2+7x−5f(x) = 5x^{3} - 7x^{2} + 7x - 5
  • x=1x = 1

Find

f ′(1)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=15x2−14x+7f '(x) = 15x^{2} - 14x + 7
  2. Substituting

    f′(1)=15(1)−14(1)+7f '(1) = 15(1) - 14(1) + 7
  3. Evaluate

    f′(1)=15−14+7=8f '(1) = 15 - 14 + 7 = 8
Answer:
f′(1)=8f '(1) = 8

Why the other options are there

  • 0 (f evaluated, not f ′)
  • 13 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 3
Derivative evaluated at a point — Test for a Minimum (2)

For f(x) = 6x³ − 5x² + 1x − 5, what is f ′(1)?

Given

  • f(x)=6x3−5x2+1x−5f(x) = 6x^{3} - 5x^{2} + 1x - 5
  • x=1x = 1

Find

f ′(1)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=18x2−10x+1f '(x) = 18x^{2} - 10x + 1
  2. Substituting

    f′(1)=18(1)−10(1)+1f '(1) = 18(1) - 10(1) + 1
  3. Evaluate

    f′(1)=18−10+1=9f '(1) = 18 - 10 + 1 = 9
Answer:
f′(1)=9f '(1) = 9

Why the other options are there

  • -3 (f evaluated, not f ′)
  • 14 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 4
Derivative evaluated at a point — Test for a Minimum (3)

For f(x) = 5x³ − 9x² + 2x − 5, what is f ′(4)?

Given

  • f(x)=5x3−9x2+2x−5f(x) = 5x^{3} - 9x^{2} + 2x - 5
  • x=4x = 4

Find

f ′(4)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=15x2−18x+2f '(x) = 15x^{2} - 18x + 2
  2. Substituting

    f′(4)=15(16)−18(4)+2f '(4) = 15(16) - 18(4) + 2
  3. Evaluate

    f′(4)=240−72+2=170f '(4) = 240 - 72 + 2 = 170
Answer:
f′(4)=170f '(4) = 170

Why the other options are there

  • 179 (f evaluated, not f ′)
  • 175 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 5
Derivative evaluated at a point — Test for a Minimum (4)

For f(x) = 2x³ − 5x² + 3x − 5, what is f ′(4)?

Given

  • f(x)=2x3−5x2+3x−5f(x) = 2x^{3} - 5x^{2} + 3x - 5
  • x=4x = 4

Find

f ′(4)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=6x2−10x+3f '(x) = 6x^{2} - 10x + 3
  2. Substituting

    f′(4)=6(16)−10(4)+3f '(4) = 6(16) - 10(4) + 3
  3. Evaluate

    f′(4)=96−40+3=59f '(4) = 96 - 40 + 3 = 59
Answer:
f′(4)=59f '(4) = 59

Why the other options are there

  • 55 (f evaluated, not f ′)
  • 64 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 6
Derivative evaluated at a point — Test for a Minimum (5)

For f(x) = 6x³ − 7x² + 2x − 5, what is f ′(3)?

Given

  • f(x)=6x3−7x2+2x−5f(x) = 6x^{3} - 7x^{2} + 2x - 5
  • x=3x = 3

Find

f ′(3)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=18x2−14x+2f '(x) = 18x^{2} - 14x + 2
  2. Substituting

    f′(3)=18(9)−14(3)+2f '(3) = 18(9) - 14(3) + 2
  3. Evaluate

    f′(3)=162−42+2=122f '(3) = 162 - 42 + 2 = 122
Answer:
f′(3)=122f '(3) = 122

Why the other options are there

  • 100 (f evaluated, not f ′)
  • 127 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 7
Derivative evaluated at a point — Test for a Minimum (6)

For f(x) = 2x³ − 8x² + 7x − 5, what is f ′(3)?

Given

  • f(x)=2x3−8x2+7x−5f(x) = 2x^{3} - 8x^{2} + 7x - 5
  • x=3x = 3

Find

f ′(3)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=6x2−16x+7f '(x) = 6x^{2} - 16x + 7
  2. Substituting

    f′(3)=6(9)−16(3)+7f '(3) = 6(9) - 16(3) + 7
  3. Evaluate

    f′(3)=54−48+7=13f '(3) = 54 - 48 + 7 = 13
Answer:
f′(3)=13f '(3) = 13

Why the other options are there

  • -2 (f evaluated, not f ′)
  • 18 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 8
Derivative evaluated at a point — Test for a Minimum (7)

For f(x) = 4x³ − 6x² + 6x − 5, what is f ′(3)?

Given

  • f(x)=4x3−6x2+6x−5f(x) = 4x^{3} - 6x^{2} + 6x - 5
  • x=3x = 3

Find

f ′(3)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=12x2−12x+6f '(x) = 12x^{2} - 12x + 6
  2. Substituting

    f′(3)=12(9)−12(3)+6f '(3) = 12(9) - 12(3) + 6
  3. Evaluate

    f′(3)=108−36+6=78f '(3) = 108 - 36 + 6 = 78
Answer:
f′(3)=78f '(3) = 78

Why the other options are there

  • 67 (f evaluated, not f ′)
  • 83 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 9
Derivative evaluated at a point — Test for a Minimum (8)

For f(x) = 6x³ − 6x² + 1x − 5, what is f ′(2)?

Given

  • f(x)=6x3−6x2+1x−5f(x) = 6x^{3} - 6x^{2} + 1x - 5
  • x=2x = 2

Find

f ′(2)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=18x2−12x+1f '(x) = 18x^{2} - 12x + 1
  2. Substituting

    f′(2)=18(4)−12(2)+1f '(2) = 18(4) - 12(2) + 1
  3. Evaluate

    f′(2)=72−24+1=49f '(2) = 72 - 24 + 1 = 49
Answer:
f′(2)=49f '(2) = 49

Why the other options are there

  • 21 (f evaluated, not f ′)
  • 54 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

Example 10
Derivative evaluated at a point — Test for a Minimum (9)

For f(x) = 5x³ − 9x² + 1x − 5, what is f ′(3)?

Given

  • f(x)=5x3−9x2+1x−5f(x) = 5x^{3} - 9x^{2} + 1x - 5
  • x=3x = 3

Find

f ′(3)

Start with the thinking

  • Differentiate term by term with the power rule.
  • Substitute only after differentiating.

Step-by-step solution

  1. Differentiate

    f′(x)=15x2−18x+1f '(x) = 15x^{2} - 18x + 1
  2. Substituting

    f′(3)=15(9)−18(3)+1f '(3) = 15(9) - 18(3) + 1
  3. Evaluate

    f′(3)=135−54+1=82f '(3) = 135 - 54 + 1 = 82
Answer:
f′(3)=82f '(3) = 82

Why the other options are there

  • 52 (f evaluated, not f ′)
  • 87 (constant differentiated incorrectly)

Reference: FE Reference Handbook — Mathematics → Test for a Minimum

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