Additional Mathematics Synthesis Guide 31: Increasing and Decreasing Functions, Derivative Sign Charts, Stationary Inflexions and Second-Derivative Decisions

Differentiate → factor → critical points → sign chart → classify → check endpoints → interpret globally.

1. Increasing and decreasing through derivative sign

Where f′(x)>0, f increases as x increases. Where f′(x)<0, f decreases.

This is stronger than looking at one gradient. It describes behaviour throughout an interval.

2. Stationary does not mean turning

A stationary point occurs where

f′(x)=0.

But the graph may be a local maximum, local minimum or stationary inflexion. Classification requires more information.

3. Sign-chart classification

• f′ changes + to −: local maximum.
• f′ changes − to +: local minimum.
• f′ keeps the same sign through a stationary point: no turning; investigate stationary inflexion.

The sign chart directly records motion of the original function.

4. Worked cubic example

Let

f(x)=x³−3x²−9x+5.

Then

f′(x)=3x²−6x−9=3(x−3)(x+1).

Stationary x-values are −1 and 3.

5. Build the sign chart

For 3(x−3)(x+1):

• x<−1: both factors negative → f′ positive.
• −1<x<3: one negative, one positive → f′ negative.
• x>3: both positive → f′ positive.

So f increases, then decreases, then increases. Therefore x=−1 is a local maximum and x=3 a local minimum.

6. Complete the stationary-point coordinates

Evaluate the original function:

f(−1)=−1−3+9+5=10.

f(3)=27−27−27+5=−22.

The stationary points are (−1,10) and (3,−22).

7. Second-derivative test

For the same function,

f″(x)=6x−6.

At x=−1, f″=−12<0, confirming local maximum. At x=3, f″=12>0, confirming local minimum.

The second derivative is a shortcut when non-zero. It is not a complete replacement for sign reasoning.

8. When f″=0, the test is inconclusive

For f(x)=x⁴, f′=4x³ and f″=12x². At x=0, f′=0 and f″=0.

Yet x=0 is a local minimum because f′ changes from negative to positive.

Therefore f″=0 does not mean “inflexion”. It means the second-derivative test did not decide.

9. Stationary inflexion

For f(x)=x³,

f′(x)=3x²≥0

with f′(0)=0. The derivative is positive on both sides, so the function keeps increasing through x=0.

Also f″(x)=6x changes sign at 0. Thus (0,0) is a stationary point of inflexion.

10. A non-stationary inflexion exists too

An inflexion is about change of concavity, not necessarily zero gradient. For example, a shifted or tilted cubic can have f″=0 while f′≠0.

Keep the concepts separate: stationary concerns first derivative; inflexion concerns concavity behaviour.

11. Factor multiplicity in f′ predicts sign change

If f′ contains a factor (x−a) to an odd power, its sign typically changes through a. If the factor appears to an even power and other factors remain non-zero, the derivative sign does not change there.

This is a fast structural guide, but a sign chart remains the safest communication.

12. Worked stationary-inflexion structure

Let f′(x)=(x−2)²(x+1).

Critical points are x=−1 and 2. At x=−1, the simple factor changes sign. At x=2, the squared factor does not.

Provided the remaining factor is positive near 2, f′ remains positive on both sides of x=2. So x=2 is stationary without a turning point.

13. Increasing/decreasing intervals from a rational derivative

Suppose

f′(x)=(x−1)/(x+2)², x≠−2.

The denominator square is positive wherever defined, so the sign of f′ is controlled by x−1.

• f′<0 for x<1, excluding x=−2.
• f′>0 for x>1.

The discontinuity x=−2 splits the domain and must appear in the interval answer.

14. Critical points include derivative failures

When analysing monotonicity, also inspect points where the derivative does not exist but the function does, and domain boundaries where behaviour can change.

For many school A-Math functions the main critical points come from f′=0, but domain structure still matters.

15. Global extrema on a closed interval

To find the greatest and least values of a differentiable function on [a,b], compare:

• all stationary points inside the interval;
• the endpoints a and b.

A local maximum need not be the global maximum on the restricted interval.

16. Worked restricted-domain decision

Let f(x)=x³−3x on [−2,2].

Then f′=3(x²−1), so stationary points are x=−1,1.

• f(−2)=−2
• f(−1)=2
• f(1)=−2
• f(2)=2

Global maximum value is 2, attained at x=−1 and 2. Global minimum is −2, attained at x=−2 and 1.

17. A sign chart can answer inequalities about f

If f is increasing on an interval, then larger x-values give larger f-values there. This can support comparisons without evaluating every value directly.

Monotonicity is therefore useful beyond sketching; it is a comparison tool.

18. Parameters can control monotonicity

Suppose f′(x)=x²+k. For f to be increasing for all real x, require

x²+k≥0

for all x. The minimum of x²+k is k at x=0, so k≥0.

If strict f′>0 everywhere were required, then k>0.

19. Parameter monotonicity by discriminant

If f′ is a quadratic ax²+bx+c and we need it always positive, use the same quadratic logic as Guide 21: a>0 and discriminant <0 for strict positivity.

This links calculus monotonicity back to quadratic parameter conditions.

20. Sketching from derivative evidence

A derivative sign chart gives rise/fall structure. Stationary-point coordinates give anchor points. The second derivative can add concavity information.

A graph sketch should reflect all of these, but it should not invent intercepts or asymptotes not established by the function.

21. Common failure patterns

• Calling every f′=0 point a maximum or minimum.
• Using f″=0 as proof of inflexion.
• Finding stationary x-values but not their y-coordinates.
• Ignoring domain breaks when writing increasing/decreasing intervals.
• Using endpoint values only after forgetting interior stationary points, or vice versa.
• Expanding a factored derivative and making sign analysis harder.
• Confusing derivative sign with function sign.
• Calling a local extremum global without checking the requested domain.

22. A reliable monotonicity routine

1. Differentiate and factor where useful.
2. Record domain exclusions and endpoints.
3. Find stationary points from f′=0.
4. Split the domain at all critical values.
5. Determine the sign of f′ on every interval.
6. Classify stationary points from sign changes or a valid second-derivative test.
7. For global extrema on a restricted interval, compare interior candidates and endpoints.
8. Return to the original function for coordinates and final interpretation.

23. Practice set

1. Find increasing/decreasing intervals of f(x)=x²−4x.
2. Classify its stationary point.
3. Find stationary points of x³−3x²−9x+5.
4. Classify them using f′ signs.
5. Classify them using f″.
6. Explain why x=0 for f=x³ is not a turning point.
7. Classify x=0 for f=x⁴.
8. If f′=(x−2)²(x+1), where can f change from increasing to decreasing?
9. For f′=(x−1)/(x+2)², state increasing intervals.
10. State decreasing intervals for the same derivative.
11. Find global extrema of f=x³−3x on [−2,2].
12. If f′=x²+k, find k for f′≥0 for all real x.
13. Find k for f′>0 for all real x.
14. If f′=x²−4x+k, find k for f to be strictly increasing everywhere.
15. Why does an even-power factor in f′ often prevent sign change?
16. Why should endpoints be checked in a closed-interval optimisation?
17. What does f″<0 at a stationary point usually imply?
18. What does f″=0 imply?
19. Give one difference between stationary and inflexion.
20. Why is a factored derivative useful?

Answers

1. f′=2x−4. Decreasing for x<2, increasing for x>2.
2. Local and global minimum at x=2.
3. x=−1,3; coordinates (−1,10),(3,−22).
4. +→− at −1 gives local maximum; −→+ at 3 gives local minimum.
5. f″=6x−6: negative at −1, positive at 3.
6. f′=3x² is positive on both sides; the function keeps increasing.
7. Minimum; f′=4x³ changes −→+.
8. At x=−1; the squared factor at x=2 does not itself change derivative sign.
9. (1,∞), excluding domain issue already to left.
10. (−∞,−2) and (−2,1).
11. Maximum 2 at x=−1,2; minimum −2 at x=−2,1.
12. k≥0.
13. k>0.
14. Quadratic f′=x²−4x+k must be >0 for all x: discriminant 16−4k<0, so k>4.
15. Its sign remains non-negative/non-positive through the repeated root if other factors do not change sign.
16. A global extremum may occur at an endpoint even when no stationary point is there.
17. Local maximum, under the standard smoothness conditions.
18. The second-derivative test is inconclusive.
19. Stationary means f′=0; inflexion concerns a change of concavity.
20. Its factors reveal critical roots and sign changes directly.

24. What mastery looks like

Mastery means the learner uses derivative sign over intervals rather than treating f′=0 as an automatic turning-point label, can distinguish stationary inflexions from extrema, knows when the second-derivative test is decisive or inconclusive, and checks endpoints for restricted global decisions.

The transfer test is to give only a factored derivative and a domain. If the learner can reconstruct where the original function rises, falls, turns or flattens without needing a graphing cue, derivative-sign reasoning is operating independently.

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Continue through Batch 08

• Guide 29: Surds, Conjugates, Rationalising Denominators, Surd Equations and Exact Verification
• Guide 30: Integration Rule Selection — Rational Powers, Exponentials, Trigonometric Forms and Reverse Chain Rule
• Guide 32: Transformed Trigonometric Graphs — Amplitude, Period, Midline, Tangent Structure and Modelling
• Additional Mathematics Directory
• BTT Mathematics Hub