Combine several pieces of derivative evidence
Three facts, one cubic
A cubic has the form \(f(x) = x^3 + ax^2 + bx + c\), where \(a\), \(b\) and \(c\) are unknown numbers. The curve shown is one cubic that could be \(f\).
Press "Add the first fact". The fact is that \(f\) has an inflection point at \(x = 1\). Only cubics that change their bend there stay, and the curve moves through some of them.
Press "Add the second fact". This inflection point is \((1, -2)\), which fixes the height at \(x = 1\), so fewer cubics stay.
Press "Add the third fact". The tangent is flat at \(x = 3\). One curve is left, because each fact removed one unknown number.
From a mark to an equation
Press each button to mark a fact on \(f\). A point gives a value of \(f\), a flat tangent gives \(f' = 0\), and a change of bend gives \(f'' = 0\). Under the last two, the faint line meets \(f'\) or \(f''\) on its axis.
Each mark gives one equation, and a formula with \(n\) unknown numbers needs \(n\) equations. With fewer marks than unknowns, many curves still fit.
Notation
Write each fact as an equation.
\[\begin{aligned} f(c) &= k \\ f'(c) &= 0 \\ f''(c) &= 0 \end{aligned}\]Count the unknown numbers. Here there are three, \(a\), \(b\) and \(c\), so you need three equations.
\[f(x) = x^3 + ax^2 + bx + c\]Solve the equations together. The first gives \(a\), which then removes one unknown from the other two.
\[\begin{aligned} f''(1) = 6 + 2a &= 0 \\ f(1) = 1 + a + b + c &= -2 \\ f'(3) = 27 + 6a + b &= 0 \end{aligned}\]The first gives \(a = -3\). Then the third becomes \(9 + b = 0\), so \(b = -9\), and the second becomes \(-11 + c = -2\), so \(c = 9\).
Test the result against every fact. The equations used only \(f''(1) = 0\), but an inflection point needs \(f''\) to change sign at 1. Its chart is red before 1 and green after, so the sign changes from negative to positive.
Under \(f'\), the chart has its zero at 3, so the tangent is flat there.
When it fails
Press "Remove the last fact", then "Try another curve that fits". With two equations for three unknowns, a whole family of curves fits.
Each curve fits every fact that remains, so a formula picked from the family is a guess, not a conclusion.
Example 1: describe f from signs
A function has these facts. \(f'\) is positive for \(x < -2\), negative for \(-2 < x < 3\), and positive for \(x > 3\). \(f''\) is negative for \(x < \tfrac12\) and positive for \(x > \tfrac12\). Describe \(f\).
At \(-2\), \(f'\) goes from positive to negative, so \(f\) has a local maximum there. At 3 it goes from negative to positive, so \(f\) has a local minimum.
At \(\tfrac12\), \(f''\) goes from negative to positive, so the bend changes from down to up and \(f\) has an inflection point there.
Combine the facts. The curve is green where \(f\) rises and red where it falls, and \(\tfrac12\) lies on the red stretch. This means \(f\) rises to the maximum at \(-2\), falls through the inflection point at \(\tfrac12\) to the minimum at 3, and rises again.
The graphs show one function that fits, \(f(x) = \tfrac{x^3}{3} - \tfrac{x^2}{2} - 6x\), with \(f' = (x+2)(x-3)\) and \(f'' = 2x - 1\).
Example 2: a parabola with a minimum
Find \(a\) and \(b\) so that \(f(x) = x^2 + ax + b\) has a local minimum at \((3, -4)\).
Two unknowns need two equations. The minimum gives two facts: the tangent is flat at \(x = 3\), and the orange ring at \((3, -4)\) must lie on the curve.
A tempting wrong move: The tangent is flat at 3, so \(f'(3) = 2(3) + a = 0\). Then \(a = -6\) and \(f(x) = x^2 - 6x\).
\[f(x) = x^2 - 6x\]This drops \(b\). The lowest point of \(x^2 - 6x\) is \((3, -9)\), in red, below the ring at \((3, -4)\). Every curve \(x^2 - 6x + b\) has a flat tangent at 3, so this fact leaves \(b\) free.
The point \((3, -4)\) gives the second equation.
\[\begin{aligned} f(3) = 9 - 18 + b &= -4 \\ b &= 5 \end{aligned}\]Test the result. The tangent at 3 is flat, and \(f''(x) = 2\) is positive, so \(f(x) = x^2 - 6x + 5\) has a minimum at 3, with \(f(3) = -4\).
Practice
Find \(a\) and \(b\) so that \(f(x) = x^3 + ax^2 + bx\) has a local minimum at \(x = 2\) and passes through \((1, -5)\).
The orange tangent at \(x = 2\) must be flat. The curve must pass through the ring. Each gives one equation.
Solution
Since \(f'(x) = 3x^2 + 2ax + b\), the flat tangent gives \(f'(2) = 0\). The point gives \(f(1) = -5\).
\[\begin{aligned} 12 + 4a + b &= 0 \\ 1 + a + b &= -5 \end{aligned}\]Subtract the second equation from the first to remove \(b\). Then \(3a = -6\), so \(a = -2\), and the second equation gives \(b = -4\).
\[11 + 3a = 5\]Test the result. \(f''(2) = 8\) is positive, so \(f(x) = x^3 - 2x^2 - 4x\) has a minimum at 2.
\[f''(x) = 6x - 4\]
Questions
The graphs show \(f\), \(f'\) and \(f''\) for one function.
Which description fits all the graphs?
- A local maximum at \(x = 2\), and an inflection point at \(x = 4\) where \(f\) is falling
- A local maximum at \(x = 2\), and an inflection point at \(x = 2\)
- A local minimum at \(x = 2\), and an inflection point at \(x = 4\)
- A local maximum at \(x = 4\), and an inflection point at \(x = 2\)
Answer
A local maximum at \(x = 2\), and an inflection point at \(x = 4\) where \(f\) is falling
What confirms that \(x = 4\) is an inflection point?
- \(f''\) is 0 at 4 and changes sign there
- \(f'\) is 0 at 4
- \(f\) is 0 at 4
- \(f'\) changes sign at 4
Answer
\(f''\) is 0 at 4 and changes sign there
The function \(f(x) = x^3 + kx^2\) has an inflection point at \(x = 2\). What is \(k\)?
- \(-6\)
- \(-3\)
- \(6\)
- \(-12\)
Answer
\(-6\)
Summary
You write each fact as an equation, solve for the unknown numbers, and test the result against every fact, signs included.
This works because a value, a flat tangent and a change of bend each give one equation, and \(n\) unknowns need \(n\) equations.
You use it when a problem gives several facts about \(f\) and asks for a formula, or for the description that fits all of them.
This is one of five skills in topic 5.9 of Which & Why: AP Calculus.