Report an inflection point

Two coordinates for one point

The top graph shows \(f(x) = x^3 - 6x^2 + 4x + 3\). The lower graph shows \(f''(x) = 6x - 12\), shaded red where \(f''\) is negative and green where it is positive.

The second derivative is 0 at \(x = 2\). It is negative to the left of 2 and positive to the right, so the concavity changes at \(x = 2\).

Move the point to \(x = 2\). The orange bar from the x-axis to the point shows the height of the graph at \(x = 2\), which is \(f(2) = -5\).

The point of inflection is \((2, -5)\). The input 2 comes from \(f''\), and the height \(-5\) comes from \(f\).

The point lies on the graph

The marked input \(c\) is an x-value where \(f''\) changes sign, so the concavity of the curve changes there.

The point of inflection is on the graph of \(f\), so its y-coordinate is \(f(c)\). The dashed orange line shows this height, and the point is \((c, f(c))\).

Move the orange point to \(c\). The tangent line crosses from one side of the curve to the other.

Notation

1. First find \(c\). Solve \(f''(c) = 0\), or find where \(f''\) does not exist, and confirm that \(f''\) changes sign at \(c\).

2. Then substitute \(c\) into \(f\), the original function. The result is the height of the graph at \(c\), shown by the dashed orange line.

   \[y = \cy{f(c)}\]

3. Write the point with its two coordinates.

   \[\bigl(c,\ \cy{f(c)}\bigr)\]

4. If the question asks for the x-value, or for where the concavity changes, give \(c\). If it asks for the point or its coordinates, give both numbers.

When it fails

The point must lie on the graph of \(f\). Press a button to read the height from a different function at \(x = 2\).

If you read the height from \(f'\), the result is \(f'(2) = -8\), and the point \((2, -8)\) is not on the graph. If you read it from \(f''\), the result is 0, and the point \((2, 0)\) is not on the graph either. Only \(f(2) = -5\) gives a point on the graph.

Example 1: A cubic

Find the point of inflection of \(g(x) = 2x^3 + 6x^2 - 3x\).

1. Differentiate twice. The second derivative is 0 at \(x = -1\).

   \[\begin{gathered} g'(x) = 6x^2 + 12x - 3 \\ g''(x) = 12x + 12 \end{gathered}\]

2. The second derivative is negative to the left of \(-1\) and positive to the right, so the concavity changes.

   \[g''(-2) = \cr{-12}, \qquad g''(0) = \cg{12}\]

3. Substitute \(x = -1\) into \(g\), the original function. This gives the height of the point.

   \[g(-1) = -2 + 6 + 3 = \cy{7}\]

4. The point of inflection is \((-1, 7)\).

Example 2: An exponential

Find the point of inflection of \(h(x) = (x - 1)e^{x}\).

1. Differentiate twice.

   \[\begin{gathered} h'(x) = xe^x \\ h''(x) = (x + 1)e^x \end{gathered}\]

2. The factor \(e^x\) is positive, so \(h''\) has the sign of \(x + 1\). It changes from negative to positive at \(x = -1\).

3. A tempting wrong move: The height of the point is \(h''(-1)\), which is 0. The point of inflection is \((-1, 0)\).

   \[\bigl(-1,\ h''(-1)\bigr) = (-1,\ 0)\]

   The point \((-1, 0)\) lies on the x-axis, and the graph meets the x-axis only at \(x = 1\). At \(x = -1\) the graph is below the axis, so \((-1, 0)\) is not on it.

4. Substitute \(x = -1\) into \(h\), the original function. The point of inflection is \(\left(-1,\ -\tfrac{2}{e}\right)\).

   \[h(-1) = (-1 - 1)e^{-1} = -\dfrac{2}{e}\]

Practice

Find the coordinates of the point of inflection of \(p(x) = \tfrac13 x^3 - 2x^2 + 3x - 1\).

Find the x-value from \(p''\), and confirm that the sign changes. Then substitute it into \(p\), not into \(p'\) or \(p''\).

Questions

These questions use the graph of \(q(x) = x^3 - 3x^2 + 4\). The top graph shows \(q\), with a hilltop at \((0, 4)\) and a valley at \((2, 0)\). The lower graph shows \(q''(x) = 6x - 6\).

1. What are the coordinates of the point of inflection of \(q\)?

   • \((1, 2)\)
   • \((1, 0)\)
   • \((2, 1)\)
   • \((1, -3)\)
   Answer

   \((1, 2)\)

2. The orange point is at the hilltop \((0, 4)\). Is it a point of inflection?

   • No. The lower graph is below the axis on both sides of \(x = 0\).
   • Yes, because the tangent at the hilltop is flat.
   • Yes, because the graph is highest there.
   Answer

   No. The lower graph is below the axis on both sides of \(x = 0\).

3. What is the point of inflection of \(r(x) = 2x^3 - 6x^2 + x\)?

   • \((1, -3)\)
   • \((1, 0)\)
   • \((1, -5)\)
   • \((-3, 1)\)
   Answer

   \((1, -3)\)

Summary

You find the x-value where \(f''\) changes sign, and you substitute it into \(f\) to find the y-coordinate.

The point of inflection is a point of the graph of \(f\), so its height is \(f(c)\), not \(f'(c)\) or \(f''(c)\).

You give both coordinates for a point, and only the x-value when the question asks where the concavity changes.

This is one of five skills in topic 5.6 of Which & Why: AP Calculus.