At the point A, the curve lies above its tangent. Which statement is correct?

y'' > 0, and the curve is concave up.

y'' < 0, and the curve is concave down.

y'' > 0, and the curve is concave down.

y'' = 0, and the curve is straight.

At the origin O, y'' = 0. What does the figure show about the concavity there?

The curve changes from concave up to concave down, because it crosses its tangent.

The curve is concave up, because y'' is not negative.

The curve is concave down, because y'' is not positive.

The curve does not change its bend, because y'' = 0.

Is the curve y^3 + 2y = x^2 + 3 concave up or concave down at (3, 2)?

Concave down, because y'' = -\tfrac{5}{343} there.

Concave up, because y' > 0 there.

Concave up, because y'' = \tfrac17 there.

The test gives no answer, because y'' = 0.

Topic 5.12

Determine concavity at a point

Builds on finding an implicit second derivative. 6 minutes.

A curve that bends both ways

The curve \(y^3 + y = x\) is hard to solve for \(y\). The orange line is the tangent at the point.

The strip between the curve and its tangent is green where \(y'' > 0\) and red where \(y'' < 0\). Drag the point to \((2, 1)\). The curve lies below its tangent there, so the strip is red.

Drag the point to the left, through the origin, where the tangent crosses the curve and the strip changes color. Past the origin the curve lies above its tangent, and the strip is green.

The sign of \(y''\) at a point tells you which way the curve bends there.

The sign of y″

Drag the point around the curve. Near it, the curve is the graph of a function \(y\) of \(x\), and the orange line is the tangent.

If \(y'' > 0\), the curve is concave up, which means it lies above its tangent, and the strip is green. If \(y'' < 0\), the curve is concave down, which means it lies below its tangent, and the strip is red.

You find \(y''\) at the point from a formula, and its sign decides the bend.

Notation

The formula for \(y''\) holds only at points of the curve. Before you use it, substitute the coordinates into the equation to confirm that the point is on the curve.
\[1^3 + 1 = 2\]
Find \(y'\) and \(y''\) by implicit differentiation. The equation gives \(3y^2y' + y' = 1\), and the quotient rule on \(y'\), with \(y'\) replaced by its formula, gives \(y''\).
\[\begin{aligned} y' &= \frac{1}{3y^2 + 1} \\ y'' &= -\frac{6y\,y'}{(3y^2 + 1)^2} \\ &= -\frac{6y}{(3y^2 + 1)^3} \end{aligned}\]
Substitute the coordinates of the point. If a formula still contains \(y'\), substitute its value at the point too.
\[y''(2, 1) = -\frac{6}{4^3} = -\frac{3}{32}\]
Read the sign. Because \(y'' < 0\), the curve is concave down at \((2, 1)\): it lies below its tangent, and the strip is red. A positive \(y''\) means concave up.
\[\cr{y'' < 0} \;\Rightarrow\; \text{concave down}\]

When it fails

The sign test needs \(y''\) to be positive or negative. Move the point to the origin, where \(y'' = 0\). The strip vanishes there, and the faint line, the tangent extended, crosses the curve.

At the origin the sign test gives no answer, because the curve is concave up on one side and concave down on the other. To see the bend near such a point, find the sign of \(y''\) on each side.

Example 1: A cubic in x

Is the curve \(y^2 = x^3 + 1\) concave up or concave down at \((2, 3)\)?

The point is on the curve, because \(3^2 = 2^3 + 1\). Differentiate both sides to find \(y'\), and evaluate it at the point.
\[\begin{aligned} 2y\,\cy{y'} &= 3x^2 \\ \cy{y'} &= \frac{3x^2}{2y} = \frac{12}{6} = \cy{2} \end{aligned}\]
Differentiate \(y'\) with the quotient rule. Because \(y\) depends on \(x\), the derivative of \(y\) is \(\cy{y'}\).
\[y'' = \frac{6x\cdot2y - 3x^2\cdot2\,\cy{y'}}{4y^2}\]
Substitute \(x = 2\), \(y = 3\) and \(y' = 2\). You do not need to simplify first.
\[y''(2, 3) = \frac{72 - 48}{36} = \cg{\tfrac23}\]
Because \(y'' > 0\), the curve is concave up at \((2, 3)\). The green strip shows that the curve lies above its tangent.

Example 2: A falling curve

Is the curve \(x^2y = 18\) concave up or concave down at \((3, 2)\)?

The point is on the curve, because \(9\cdot2 = 18\). Differentiate both sides to find \(y'\), and evaluate it at the point.
\[\begin{aligned} 2xy + x^2\,\cy{y'} &= 0 \\ \cy{y'} &= -\frac{2y}{x} = -\frac43 \end{aligned}\]
A tempting wrong move: The slope \(-\tfrac43\) is negative, so the curve is concave down.
\[y' < 0 \;\Rightarrow\; \text{concave down}\]
The slope tells you that the curve falls, not how it bends. In the figure the curve falls at \((3, 2)\), but it lies above its tangent, and the strip is green.
Concavity depends on \(y''\). Differentiate \(y' = -\tfrac{2y}{x}\) with the quotient rule, and substitute the point and \(y' = -\tfrac43\).
\[\begin{aligned} y'' &= -\frac{2\,(x\,y' - y)}{x^2} \\ &= -\frac{2\,(-4 - 2)}{9} = \cg{\tfrac43} \end{aligned}\]
Because \(y'' > 0\), the curve is concave up at \((3, 2)\).

Practice

Is the curve \(y + \ln y = x\) concave up or concave down at \((1, 1)\)? The domain is \(y > 0\).

Find \(y'\) first. Then differentiate \(y'\) and substitute the point. Drag the point to \((1, 1)\) and compare the color of the strip with your sign.

Solution

The point is on the curve, because \(1 + \ln 1 = 1\). Differentiate both sides to find \(y'\).
\[\begin{aligned} y' + \frac{y'}{y} &= 1 \\ y'(y + 1) &= y \\ \cy{y'} &= \frac{y}{y + 1} \end{aligned}\]
Differentiate \(\cy{y'}\) with the quotient rule. The two terms \(y\,\cy{y'}\) cancel.
\[y'' = \frac{\cy{y'}(y + 1) - y\,\cy{y'}}{(y + 1)^2} = \frac{\cy{y'}}{(y + 1)^2}\]
At \((1, 1)\), the value of \(y'\) is \(\tfrac12\). Substitute it.
\[y''(1, 1) = \frac{1/2}{4} = \cg{\tfrac18}\]
Because \(y'' > 0\), the curve is concave up at \((1, 1)\), and the strip is green.

Questions

The graph shows the curve \(y^3 + y = x\), where \(y'' = -\tfrac{6y}{(3y^2 + 1)^3}\). The orange point is \(A = (-2, -1)\). The faint line is the tangent at the origin \(O\).

At the point \(A\), the curve lies above its tangent. Which statement is correct?
- \(y'' > 0\), and the curve is concave up.
- \(y'' < 0\), and the curve is concave down.
- \(y'' > 0\), and the curve is concave down.
- \(y'' = 0\), and the curve is straight.
Answer
\(y'' > 0\), and the curve is concave up.
At the origin \(O\), \(y'' = 0\). What does the figure show about the concavity there?
- The curve changes from concave up to concave down, because it crosses its tangent.
- The curve is concave up, because \(y''\) is not negative.
- The curve is concave down, because \(y''\) is not positive.
- The curve does not change its bend, because \(y'' = 0\).
Answer
The curve changes from concave up to concave down, because it crosses its tangent.
Is the curve \(y^3 + 2y = x^2 + 3\) concave up or concave down at \((3, 2)\)?
- Concave down, because \(y'' = -\tfrac{5}{343}\) there.
- Concave up, because \(y' > 0\) there.
- Concave up, because \(y'' = \tfrac17\) there.
- The test gives no answer, because \(y'' = 0\).
Answer
Concave down, because \(y'' = -\tfrac{5}{343}\) there.

Summary

You check that the point is on the curve, find \(y''\), substitute the point, and read the sign.

This works because the curve lies above its tangent where \(y'' > 0\), and below it where \(y'' < 0\).

You use it to find the bend of a curve whose equation cannot be solved for \(y\).

This is one of four skills in topic 5.12 of Which & Why: AP Calculus.