Analysing a curve with its derivative
The derivative gives the stationary inputs and the direction in which the function moves. Evaluate the original function at the resulting inputs to obtain the points on the curve
\[f'(x)=0\text{ at a stationary input},\quad\text{example: }f(x)=x^{3}-3x\]
- Find stationary inputs A stationary input is found by solving the derivative equation equal to zero \(\displaystyle f'(x)=0\quad\Rightarrow\quad x=a\)
- Read the sign change A negative-to-positive derivative change gives a local minimum, while a positive-to-negative change gives a local maximum \(\displaystyle -\;\to\;+\;\Rightarrow\;\text{minimum},\quad +\;\to\;-\;\Rightarrow\;\text{maximum}\)
- Use derivative signs for intervals If the derivative is positive throughout an interval, the function is increasing there. If it is negative throughout an interval, the function is decreasing there \(\displaystyle f'(x)>0\text{ for all }x\in I\Rightarrow f\text{ increasing on }I,\quad f'(x)<0\text{ for all }x\in I\Rightarrow f\text{ decreasing on }I\)
- Evaluate the original function Use the original function to find the ordinate of each stationary point and any endpoint \(\displaystyle f(-1)=(-1)^{3}-3(-1)=2\)
Worked example
Find and classify the stationary points of the function
\[g(x)=x^{3}-6x^{2}+9x\]
- \(\displaystyle g'(x)=3x^{2}-12x+9=3(x-1)(x-3)\)Differentiate the polynomial and factorise the derivative
- \(\displaystyle g'(x)=0\quad\Rightarrow\quad x=1\text{ or }x=3\)The stationary inputs are the solutions of the derivative equation
- \(\displaystyle g'(0)=9>0,\quad g'(2)=-3<0,\quad g'(4)=9>0\)The signs change positive to negative at 1 and negative to positive at 3.
- \(\displaystyle g(1)=4,\quad g(3)=0\)The stationary points are a local maximum at (1,4) and a local minimum at (3,0)
Common mistakes
Solving the wrong equation
✕\(\displaystyle f(x)=0\quad\Rightarrow\quad x=-\sqrt{3},0,\sqrt{3}\)✓\(\displaystyle f'(x)=0\quad\Rightarrow\quad 3x^{2}-3=0\quad\Rightarrow\quad x=-1,1\)
Stationary inputs come from f'(x)=0, while f(x)=0 finds x-interceptsReversing derivative signs
✕\(\displaystyle f'(x)>0\quad\Rightarrow\quad f\text{ is decreasing}\)✓\(\displaystyle f'(x)>0\text{ for all }x\in I\quad\Rightarrow\quad f\text{ is increasing on }I\)
A positive derivative throughout an interval means the function rises as the input increases across that intervalUsing the derivative for the ordinate
✕\(\displaystyle f'(-1)=0\quad\Rightarrow\quad y=0\)✓\(\displaystyle f(-1)=(-1)^{3}-3(-1)=2\quad\Rightarrow\quad y=2\)
The derivative gives the stationary condition, while the original function gives the y-coordinateDifficulty levels
introAnalyse a cubic using two stationary inputs, a sign table and function features
standardUse several derivative sign intervals for stationary, nature or increasing-decreasing tasks
challengeCompare stationary and endpoint values on a restricted domain or select a consistent curve
examComplete a multi-part feature analysis before selecting an interval, graph or extrema conclusion
Questions about this topic
How are stationary inputs found?+
Differentiate the function and solve f'(x) = 0. The solutions are inputs where the tangent is horizontal
How is a stationary point classified?+
Check the derivative signs on either side. Negative to positive gives a local minimum; positive to negative gives a local maximum. No sign change does not by itself prove an inflection: check that the concavity changes there
Why is the original function used after solving f'(x) = 0?+
The derivative supplies the x-coordinates, but only the original function supplies the corresponding y-coordinates
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 6: Applications of differentiation
In this chapter
Differentiate polynomialsChallenge enrichment: rational powers and derivative domainsDifferentiate natural exponential functions and solve simple derivative conditionsDifferentiate sine, cosine and tangent functions and solve simple derivative conditionsDifferentiate a composite function by the chain ruleDifferentiate a product or quotient of two functionsEvaluate and determine function rules6B Differentiation rules: Find the maximal domain of an algebraic functionExpand and factor polynomial expressionsSimplify and combine algebraic fractions6B Differentiation rules: Find the equation of a line from its graph or from its intercept and slopeAnalyse polynomial curves, stationary points and extremaUse a derivative to analyse and recognise the graph of a function6C Tangents, normals and curve sketching: Evaluate a polynomial or recover a coefficient parameter6C Tangents, normals and curve sketching: Move between roots, factors, polynomial rules and equationsRecognise and classify a polynomialConnect a factorised higher-degree rule and its graphAnalyse a cubic graph or determine its ruleWork with sets and interval notation6C Tangents, normals and curve sketching: Find the maximal domain of an algebraic function6C Tangents, normals and curve sketching: Find the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular lineSolve a maximum or minimum problem by differentiationSet up and solve a practical linear equationRearrange literal equations and recognise equivalent line forms6D Optimisation: Find the maximal domain of an algebraic function6D Optimisation: Evaluate a polynomial or recover a coefficient parameter6D Optimisation: Move between roots, factors, polynomial rules and equationsCalculate and interpret average and instantaneous rates from a modelVolume of cones, pyramids and spheresVolume of prisms and cylindersPerimeter and area of standard shapesSimilar figures and scale factorsFind a side of a right-angled triangleAnalyse straight-line motion through displacement, velocity and acceleration6F Kinematics: Evaluate a polynomial or recover a coefficient parameter