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Applications of differentiation

Classify stationary points · Diagrams · Complete reading

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Increasing and decreasing intervals

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Illustrative source-labelled curve through A,B,C,D: rising from a to b, falling from b to c, rising from c to d. B is a local peak and C a local trough.
Illustrative source-labelled curve through A,B,C,D: rising from a to b, falling from b to c, rising from c to d. B is a local peak and C a local trough.
Illustrative source-labelled curve through A,B,C,D: rising from a to b, falling from b to c, rising from c to d. B is a local peak and C a local trough.
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Method map: Classify a stationary point

Start at an interior input a where f′(a) = 0. Either test can establish a local maximum or minimum; an inconclusive test needs more information.

  • First-derivative test

    Use signs that hold on small intervals on either side of a. Two isolated sample values are not a proof.

    f′(a)=0f^{\prime}(a)=0

    Positive → zero → negativeLocal maximum

    The curve rises into the point, then falls. Its value is greater than nearby values.

    Negative → zero → positiveLocal minimum

    The curve falls into the point, then rises. Its value is less than nearby values.

    Positive on both sides, or negative on both sidesNo local extremum

    There is no turning point. Confirm a change of concavity before calling this a stationary inflection; the same-sign pattern alone does not prove one.

    These conclusions are local, not a comparison with the whole graph. If the one-sided signs do not settle, this chart is inconclusive; use another local argument.

  • Second-derivative test

    At the same stationary input, apply the test under the usual local differentiability assumptions. This is an alternative test; you do not have to perform both tests.

    f′(a)=0f^{\prime}(a)=0

    Negative second derivativeLocal maximum

    f′′(a)<0f^{\prime\prime}(a)<0 establishes a strict local maximum.

    Positive second derivativeLocal minimum

    f′′(a)>0f^{\prime\prime}(a)>0 establishes a strict local minimum.

    Zero or unavailable second derivativeInconclusive

    Return to the first-derivative test or another valid local argument. A zero second derivative does not prove an inflection.

    The worked comparisons below show three different outcomes with f′′(0)=0f^{\prime\prime}(0)=0. Predict each outcome before opening its reasoning.

    Find the point’s height by substituting a into f, not into a derivative. For an optimisation problem, compare valid candidates with relevant domain boundaries too.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Compare the cubic at zero

For the illustrative function f(x)=x3f(x)=x^3, classify the stationary point at x=0x=0. All three comparisons have f′′(0)=0f^{\prime\prime}(0)=0: what extra evidence decides this one?

Show reasoning
  1. 1. Recognise the inconclusive test

    The input is stationary, but a zero second derivative does not classify the point.

    f′(x)=3x2,f′′(x)=6x,f′(0)=f′′(0)=0.\begin{aligned}f^{\prime}(x)&=3x^2,\\f^{\prime\prime}(x)&=6x,\\f^{\prime}(0)&=f^{\prime\prime}(0)=0.\end{aligned}
  2. 2. Check the signs on intervals

    For every x<0x<0 and every x>0x>0, 3x2>03x^2>0. The curve is increasing on both sides, so the point is neither a local maximum nor a local minimum. These are interval statements, not just two sampled slopes.

  3. 3. Establish the change of concavity

    The second derivative 6x6x is negative to the left and positive to the right of zero. Concavity changes, and f(0)=0f(0)=0, so (0,0)(0,0) is a stationary point of inflection. Same-sign first derivatives alone would not establish the concavity change.

Compare the quartic at zero

For the illustrative function f(x)=x4f(x)=x^4, classify the stationary point at x=0x=0. Does f′′(0)=0f^{\prime\prime}(0)=0 force the same outcome as the cubic? Predict the signs of f′f^{\prime} on either side.

Show reasoning
  1. 1. Recognise the same inconclusive result

    Both derivatives vanish at zero. The second-derivative test supplies no classification.

    f′(x)=4x3,f′′(x)=12x2,f′(0)=f′′(0)=0.\begin{aligned}f^{\prime}(x)&=4x^3,\\f^{\prime\prime}(x)&=12x^2,\\f^{\prime}(0)&=f^{\prime\prime}(0)=0.\end{aligned}
  2. 2. Use the first-derivative sign change

    For every x<0x<0, 4x3<04x^3<0; for every x>0x>0, 4x3>04x^3>0. The curve falls into the point and rises away, establishing a strict local minimum at (0,0)(0,0). The height comes from f(0)f(0), not from a derivative.

  3. 3. Check with the function values

    Every nonzero nearby input has x4>0=f(0)x^4>0=f(0). This independently confirms the strict minimum. Also, 12x2>012x^2>0 on both sides: concavity does not change at zero, so this is not an inflection.

Reflect the quartic in the x-axis

For the illustrative function f(x)=−x4f(x)=-x^4, classify the stationary point at x=0x=0. The second derivative is still zero there. Predict what reflecting the quartic does to its first-derivative signs.

Show reasoning
  1. 1. Keep the test's limitation

    The negative sign changes the nearby slopes, but both derivatives at zero are still zero.

    f′(x)=−4x3,f′′(x)=−12x2,f′(0)=f′′(0)=0.\begin{aligned}f^{\prime}(x)&=-4x^3,\\f^{\prime\prime}(x)&=-12x^2,\\f^{\prime}(0)&=f^{\prime\prime}(0)=0.\end{aligned}
  2. 2. Reverse the sign change

    For every x<0x<0, −4x3>0-4x^3>0; for every x>0x>0, −4x3<0-4x^3<0. The curve rises into the point and falls away, establishing a strict local maximum at (0,0)(0,0).

  3. 3. Check the value and concavity

    Every nonzero nearby input has −x4<0=f(0)-x^4<0=f(0). The second derivative is negative on both sides, so concavity does not change. Across the three comparisons, the same zero second derivative permits an inflection, a minimum or a maximum.

Compare with the curve sketches

Compare the four local configurations below. All four functions are defined for every real x; each plot shows only the stated window. These functions illustrate the source’s four cases, not equations inferred from its combined A/B/C sketch.

Local maximum: rising, then falling

f(x)=x3−3xf(x)=x^3-3x. Inspect −2≤x≤0-2\le x\le0 around the stationary point (−1,2)(-1,2).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On −2<x<−1-2<x<-1, f′(x)>0f^{\prime}(x)>0: rising.

At the point

f′(−1)=0f^{\prime}(-1)=0. Horizontal tangent y=2y=2.

Right of the point

On −1<x<0-1<x<0, f′(x)<0f^{\prime}(x)<0: falling.

Since f′(x)=3(x−1)(x+1)f^{\prime}(x)=3(x-1)(x+1), these signs hold throughout the stated intervals. They establish a strict local maximum at (−1,2)(-1,2). Also f′′(−1)=−6<0f^{\prime\prime}(-1)=-6<0. The full cubic is unbounded above, so this is not a whole-graph maximum.

Local minimum: falling, then rising

f(x)=x3−3xf(x)=x^3-3x. Inspect 0≤x≤20\le x\le2 around the stationary point (1,−2)(1,-2).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On 0<x<10<x<1, f′(x)<0f^{\prime}(x)<0: falling.

At the point

f′(1)=0f^{\prime}(1)=0. Horizontal tangent y=−2y=-2.

Right of the point

On 1<x<21<x<2, f′(x)>0f^{\prime}(x)>0: rising.

The factorisation f′(x)=3(x−1)(x+1)f^{\prime}(x)=3(x-1)(x+1) establishes these signs on whole intervals, giving a strict local minimum at (1,−2)(1,-2). Also f′′(1)=6>0f^{\prime\prime}(1)=6>0. The full cubic is unbounded below, so this is not a whole-graph minimum.

Stationary inflection: rising on both sides

f(x)=x3f(x)=x^3. Inspect −1≤x≤1-1\le x\le1 around (0,0)(0,0).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On −1<x<0-1<x<0, f′(x)=3x2>0f^{\prime}(x)=3x^2>0.

At the point

f′(0)=0f^{\prime}(0)=0. Horizontal tangent y=0y=0.

Right of the point

On 0<x<10<x<1, f′(x)=3x2>0f^{\prime}(x)=3x^2>0.

There is no turning point. To establish an inflection as well, check f′′(x)=6xf^{\prime\prime}(x)=6x: it is negative to the left and positive to the right, so concavity changes. The zero value f′′(0)=0f^{\prime\prime}(0)=0 alone is inconclusive.

Stationary inflection: falling on both sides

f(x)=−x3f(x)=-x^3. Inspect −1≤x≤1-1\le x\le1 around (0,0)(0,0).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On −1<x<0-1<x<0, f′(x)=−3x2<0f^{\prime}(x)=-3x^2<0.

At the point

f′(0)=0f^{\prime}(0)=0. Horizontal tangent y=0y=0.

Right of the point

On 0<x<10<x<1, f′(x)=−3x2<0f^{\prime}(x)=-3x^2<0.

The curve keeps falling, so there is no turning point. Here f′′(x)=−6xf^{\prime\prime}(x)=-6x changes from positive to negative, establishing the concavity change. Same-sign first derivatives alone do not prove every function has an inflection.

A local maximum or minimum compares nearby function values. Check relevant domain boundaries separately for optimisation; the displayed window does not establish a whole-graph extremum.

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Combined illustrative curve has a local maximum A at a, a local minimum B at b and a stationary inflection C at c. Four separate panels show first-derivative signs: maximum (+, 0, −); minimum (−, 0, +); rising inflection (+, 0, +); falling inflection (−, 0, −), with horizontal tangent segments and concavity change at both inflections.
Combined illustrative curve has a local maximum A at a, a local minimum B at b and a stationary inflection C at c. Four separate panels show first-derivative signs: maximum (+, 0, −); minimum (−, 0, +); rising inflection (+, 0, +); falling inflection (−, 0, −), with horizontal tangent segments and concavity change at both inflections.
Combined illustrative curve has a local maximum A at a, a local minimum B at b and a stationary inflection C at c. Four separate panels show first-derivative signs: maximum (+, 0, −); minimum (−, 0, +); rising inflection (+, 0, +); falling inflection (−, 0, −), with horizontal tangent segments and concavity change at both inflections.
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