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Recover a polynomial rule from derivative and tangent conditions

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Recovering a function from its derivative

A derivative determines the non-constant part of a polynomial. A function value fixes the additive constant; a tangent gradient alone cannot. A known tangent line y = mx + b at x = a also supplies the function value f(a) = ma + b.

\[f'(x)=2ax+b\quad\Rightarrow\quad f(x)=ax^2+bx+c\]
  1. 1Count the unknown constants A quadratic recovered from its derivative has a constant term that differentiation cannot determine \(\displaystyle f'(x)=8x-6\quad\Rightarrow\quad f(x)=4x^2-6x+c\)
  2. 2Reverse the power rule Increase each power by one and divide the coefficient by the new power when recovering a polynomial \(\displaystyle 6x^2-4x+2\quad\Rightarrow\quad 2x^3-2x^2+2x+c\)
  3. 3Translate a tangent condition A tangent gradient gives the derivative condition f′(a) = m. If the full tangent line y = mx + b and contact input a are known, also use the point condition f(a) = ma + b. \(\displaystyle \begin{aligned}f'(a)&=m\\f(a)&=ma+b\end{aligned}\)
  4. 4Use the point condition Substitute the given point into the recovered function to find each remaining constant, then differentiate to check \(\displaystyle f(2)=9\quad\Rightarrow\quad 4(2)^2-6(2)+c=9\)
Year 12, VCE Units 3 & 4, Chapter 5, 4 difficulty levels, No calculatorBuild a worksheet

Recovering a function from its derivative

A derivative determines the non-constant part of a polynomial. A function value fixes the additive constant; a tangent gradient alone cannot. A known tangent line y = mx + b at x = a also supplies the function value f(a) = ma + b.

\[f'(x)=2ax+b\quad\Rightarrow\quad f(x)=ax^2+bx+c\]
  1. Count the unknown constants A quadratic recovered from its derivative has a constant term that differentiation cannot determine \(\displaystyle f'(x)=8x-6\quad\Rightarrow\quad f(x)=4x^2-6x+c\)
  2. Reverse the power rule Increase each power by one and divide the coefficient by the new power when recovering a polynomial \(\displaystyle 6x^2-4x+2\quad\Rightarrow\quad 2x^3-2x^2+2x+c\)
  3. Translate a tangent condition A tangent gradient gives the derivative condition f′(a) = m. If the full tangent line y = mx + b and contact input a are known, also use the point condition f(a) = ma + b. \(\displaystyle \begin{aligned}f'(a)&=m\\f(a)&=ma+b\end{aligned}\)
  4. Use the point condition Substitute the given point into the recovered function to find each remaining constant, then differentiate to check \(\displaystyle f(2)=9\quad\Rightarrow\quad 4(2)^2-6(2)+c=9\)

Worked example

A polynomial g has derivative g'(x)=6x^2-4x+2 and passes through the point (1,5). Find g(2)

\[g'(x)=6x^2-4x+2,\quad g(1)=5\]
  1. \(\displaystyle g(x)=2x^3-2x^2+2x+c\)Reverse each power-rule term and include the unknown constant c
  2. \(\displaystyle g(1)=2(1)^3-2(1)^2+2(1)+c=5\quad\Rightarrow\quad c=3\)Substitute the given point and solve the resulting equation
  3. \(\displaystyle g(x)=2x^3-2x^2+2x+3\)Replace c with 3 in the recovered polynomial
  4. \(\displaystyle g(2)=2(2)^3-2(2)^2+2(2)+3=16-8+4+3=15\)Substitute x = 2 to obtain the required function value

Common mistakes

Omitting the constant
✕\(\displaystyle f'(x)=8x-6\quad\Rightarrow\quad f(x)=4x^2-6x\)✓\(\displaystyle f'(x)=8x-6\quad\Rightarrow\quad f(x)=4x^2-6x+c\)
Differentiation removes a constant, so the recovered function must include an unknown constant until a point condition determines it
Not reducing the exponent
✕\(\displaystyle \frac{d}{dx}(3x^2)=6x^2\)✓\(\displaystyle \frac{d}{dx}(3x^2)=6x\)
Differentiation multiplies by the exponent and reduces the exponent by one
Using a point value as a derivative value
✕\(\displaystyle f(1)=5\quad\Rightarrow\quad f'(1)=5\)✓\(\displaystyle f(x)=3x^2-4x+c,\quad f(1)=5\quad\Rightarrow\quad 3-4+c=5\)
A point condition gives a value of the original function, so it must be substituted into f rather than f'

Difficulty levels

introRecover a quadratic from its derivative and one function value
standardRecover a cubic from its derivative and two point conditions
challengeUse a tangent-line condition and a point condition to recover a quadratic
examRecover a cubic from derivative and tangent conditions, then use the result in a gradient or value calculation

Questions about this topic

Why is an arbitrary constant needed when reversing a derivative?+

All constant terms have derivative zero, so the derivative alone cannot identify which constant was present

How is a tangent condition used?+

A tangent gradient gives the derivative at the contact input, but does not determine an additive constant. If the tangent line and contact input are known, the line also gives the function value there; use both conditions with the other data

How can a recovered function be checked?+

Differentiate it to recover the given derivative and substitute each given point into the function

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 5: Differentiation

In this chapter

Evaluate limits and test continuityGradient of a tangent from first principlesDifferentiate power functions and evaluate a derivativeExpand and factor polynomial expressionsPascal rows and binomial coefficient structure5A Limits, first principles and the power rule: Simplify and combine algebraic fractionsEvaluate a polynomial or recover a coefficient parameterSubstitute values into a formula5A Limits, first principles and the power rule: Solve a linear equationTest or construct a smooth piecewise joinRead and construct a gradient function from a function graphAnalyse a cubic graph or determine its ruleEvaluate and determine function rulesEquation of a tangent or normal lineRecover a polynomial rule from derivative and tangent conditionsFind the gradient of a line through two pointsFind the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular lineSolve a pair of simultaneous linear equationsDetermine parameters in a polynomial identity5C Tangents, normals and derivative conditions: Solve a linear equationDifferentiate natural exponential functions and solve simple derivative conditionsExponential growth and decay rate models5D Exponential derivatives and rate models: Simplify and combine algebraic fractions5D Exponential derivatives and rate models: Round to decimal places and significant figuresDifferentiate sine, cosine and tangent functions and solve simple derivative conditionsGeometric optimisation with a trigonometric modelSinusoidal models and their rates of changeSolve a practical right-angled-triangle problemPerimeter and area of standard shapes5E Trigonometric differentiation and rates: Round to decimal places and significant figuresApproximate a cubic root using Newton's method