riaology
Unit 3 · Lesson 2 of 7

Find critical points, then inspect the interval.

Choosing a maximum response requires more than finding where a derivative is zero. Endpoints may win.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its function value.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase input x? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Read the modelf(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

Input x versus f(x). Function. Selected function value: -2 . At x=−1 the derivative changes positive to negative: local maximum. At x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.-2-20022f(x)Input x
Function

function valuef′(1)=0,f′′(1)=6f'(1)=0,\quad f''(1)=6

At x=−1x=-1 the derivative changes positive to negative: local maximum. At x=1x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.

Assumptions, units, and limits

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

For an interior differentiable extremum the derivative is zero, but this is only a candidate condition. Evaluate critical points and endpoints for absolute extrema on a closed interval. Derivative sign changes distinguish local maxima and minima.

f(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x
02

Keep the reasoning visible

f′(x)=0gives candidates, not automatic extremaf'(x)=0\quad\text{gives candidates, not automatic extrema}

At x=−1x=-1 the derivative changes positive to negative: local maximum. At x=1x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.

03

What makes this model useful?

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

A stationary point may be neither a maximum nor a minimum. Non-differentiable points also deserve inspection.

A worked example

Calculate the starting scenario

  1. Use input x = 1 .
  2. f′(1)=0,f′′(1)=6f'(1)=0,\quad f''(1)=6
  3. At x=−1x=-1 the derivative changes positive to negative: local maximum. At x=1x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here. The checked quantity is function value.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the function value using input x 1 . State the meaning of its sign and units.

Checked quantity: function value · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

For an interior differentiable extremum the derivative is zero, but this is only a candidate condition. Evaluate critical points and endpoints for absolute extrema on a closed interval. Derivative sign changes distinguish local maxima and minima.

Explore the visual solution
Read the modelf(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

Input x versus f(x). Function. Selected function value: -2 . At x=−1 the derivative changes positive to negative: local maximum. At x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.-2-20022f(x)Input x
Function

function valuef′(1)=0,f′′(1)=6f'(1)=0,\quad f''(1)=6

At x=−1x=-1 the derivative changes positive to negative: local maximum. At x=1x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.

Assumptions, units, and limits

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

For an interior differentiable extremum the derivative is zero, but this is only a candidate condition. Evaluate critical points and endpoints for absolute extrema on a closed interval. Derivative sign changes distinguish local maxima and minima.

f(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x
Calculate and explain

Apply the idea to a new scenario

At x=−1x=-1 for x3−3x^3-3x, evaluate the function and classify the local behavior.

Checked quantity: function value · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

For an interior differentiable extremum the derivative is zero, but this is only a candidate condition. Evaluate critical points and endpoints for absolute extrema on a closed interval. Derivative sign changes distinguish local maxima and minima.

Explore the visual solution
Read the modelf(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

Input x versus f(x). Function. Selected function value: 2 . At x=−1 the derivative changes positive to negative: local maximum. At x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.-2-20022f(x)Input x
Function

function valuef′(−1)=0,f′′(−1)=−6f'(-1)=0,\quad f''(-1)=-6

At x=−1x=-1 the derivative changes positive to negative: local maximum. At x=1x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.

Assumptions, units, and limits

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

For an interior differentiable extremum the derivative is zero, but this is only a candidate condition. Evaluate critical points and endpoints for absolute extrema on a closed interval. Derivative sign changes distinguish local maxima and minima.

f(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelf(x)=x3−3x,f′=3(x2−1),f′′=6xf(x)=x^3-3x,\quad f'=3(x^2-1),\quad f''=6x

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

Input x versus f(x). Function. Selected function value: -2 . At x=−1 the derivative changes positive to negative: local maximum. At x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.-2-20022f(x)Input x
Function

function valuef′(1)=0,f′′(1)=6f'(1)=0,\quad f''(1)=6

At x=−1x=-1 the derivative changes positive to negative: local maximum. At x=1x=1 it changes negative to positive: local minimum. Endpoints tie the absolute values here.

Assumptions, units, and limits

Closed interval [−2,2]. Critical points and endpoints all matter for absolute extrema.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. For an interior differentiable extremum the derivative is zero, but this is only a candidate condition. Evaluate critical points and endpoints for absolute extrema on a closed interval. Derivative sign changes distinguish local maxima and minima.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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