Connect signs with direction and concavity

A population that keeps rising

The top graph shows a population \(P\), in thousands, after \(t\) years. The middle graph shows \(P'\), the rate of growth, and the bottom graph shows \(P''\).

Drag the point. The curve \(P\) is green because it rises at every \(t\), and so \(P'\) stays above the axis.

The curve \(P'\) is green where it rises and red where it falls. Before \(t = 0\), \(P'\) rises, so the tangent of \(P\) becomes steeper and \(P\) bends up.

After \(t = 0\), \(P'\) falls, so the tangent becomes flatter and \(P\) bends down. The population continues to rise, but more slowly.

Any curve

Take any curve \(f\). Press "Color f by direction". The curve is green where it rises and red where it falls, and it rises where \(f'\) is above the axis.

Press "Color f′ by bend". The graph of \(f'\) is green where it rises and red where it falls. Where \(f'\) rises, \(f''\) is above the axis and \(f\) bends up.

Each stretch of the curve has one direction and one bend, which the charts under f′ and f″ record. This curve shows all four combinations.

Drag the point. The tangent lies below the curve where it bends up and above the curve where it bends down.

Notation

1. On an interval where \(f'\) is positive, \(f\) increases, and the curve is green. Where \(f'\) is negative, \(f\) decreases, and the curve is red.

   \[\begin{array}{l} \cg{f' > 0} \;\Rightarrow\; f \text{ increases} \\ \cr{f' < 0} \;\Rightarrow\; f \text{ decreases} \end{array}\]

2. On an interval where \(f''\) is positive, \(f'\) increases, so the slope of \(f\) grows and the curve bends up. Then \(f\) is concave up and lies above its tangents.

   \[\cg{f'' > 0} \;\Rightarrow\; f \text{ is concave up}\]

3. Where \(f''\) is negative, \(f'\) decreases, so the curve bends down. Then \(f\) is concave down and lies below its tangents.

   \[\cr{f'' < 0} \;\Rightarrow\; f \text{ is concave down}\]

4. The two signs answer two questions. The sign of \(f'\) gives the direction, and the sign of \(f''\) gives the bend.

   Both tests need an interval on which \(f\) is defined and has these derivatives.

When it fails

For \(f(x) = -\frac1x\), \(f'(x) = \frac{1}{x^2}\) is positive at every \(x\) except 0, so the curve is green on both sides. Press "Move the right point past 0". The secant from \(x = -1\) to \(x = 1\) slopes down, because \(f(-1) = 1\) is greater than \(f(1) = -1\): \(f\) does not increase across 0.

The graph has a gap at \(x = 0\), where \(f\) is not defined. The test needs one interval, and each side of the gap is a separate interval.

Example 1: four stretches of one curve

Describe \(f(x) = 2x^3 - 9x^2 + 12x\) on each interval where its direction and bend do not change.

1. The sign of \(f'\) can change only where \(f' = 0\), at \(x = 1\) and \(x = 2\). A test value in each interval shows that \(f'\) is positive left of 1, negative between 1 and 2, and positive right of 2.

   \[\begin{aligned} f'(x) &= 6x^2 - 18x + 12 \\ &= 6(x - 1)(x - 2) \end{aligned}\]

2. The sign of \(f''\) can change only where \(f'' = 0\), which is at \(x = \tfrac32\). It is negative to the left and positive to the right.

   \[f''(x) = 12x - 18\]

3. Combine the two charts. For \(x < 1\), \(f\) increases and is concave down. For \(1 < x < \tfrac32\), it decreases and is concave down.

4. For \(\tfrac32 < x < 2\), \(f\) decreases and is concave up. For \(x > 2\), it increases and is concave up.

Example 2: a tank that fills

A tank fills. The depth in centimeters after \(t\) minutes is \(h(t) = 6\sqrt{t+1}\). At \(t = 3\), \(h'(3) = \tfrac32\) and \(h''(3) = -\tfrac{3}{16}\). Is the depth rising or falling at \(t = 3\)?

1. The statement gives \(h'(3)\) and \(h''(3)\), and each answers a different question.

2. A tempting wrong move: \(h''(3)\) is negative, so the depth is falling.

   \[h''(3) < 0 \Rightarrow h \text{ falls}\]

   The tangent at \(t = 3\) slopes up, so the depth rises. A negative \(h''\) means that the slope is shrinking, not that the depth drops.

3. Direction comes from \(h'\). Here \(h'(3) = \tfrac32\) is positive, so the depth is rising.

   \[h'(t) = \frac{3}{\sqrt{t+1}}\]

4. The bend comes from \(h''\). Here \(h''(3) = -\tfrac{3}{16}\) is negative, so the slope is shrinking and the curve bends down. The depth is rising, but more slowly each minute.

Practice

Let \(f'(x) = x^2 - 6x + 5\). At \(x = 2\), is \(f\) increasing or decreasing, and is it concave up or concave down?

Read the sign of \(f'(2)\) for the direction, and the sign of its derivative at 2 for the bend.

Questions

The graphs show \(f\), \(f'\) and \(f''\) for one function.

1. On the interval \(1 < x < 2\), which describes \(f\)?

   • Decreasing and concave down
   • Decreasing and concave up
   • Increasing and concave down
   • Increasing and concave up
   Answer

   Decreasing and concave down

2. At \(x = 3\), which pair of signs is correct?

   • \(f' < 0\) and \(f'' > 0\)
   • \(f' < 0\) and \(f'' < 0\)
   • \(f' > 0\) and \(f'' > 0\)
   • \(f' > 0\) and \(f'' < 0\)
   Answer

   \(f' < 0\) and \(f'' > 0\)

3. Let \(f(x) = x^3 - 12x + 5\). Which describes \(f\) at \(x = 1\)?

   • Decreasing and concave up
   • Decreasing and concave down
   • Increasing and concave up
   • Increasing and concave down
   Answer

   Decreasing and concave up

Summary

You read the sign of \(f'\) for direction and the sign of \(f''\) for bend, each on its own interval.

This works because \(f'\) is the slope of \(f\) and \(f''\) is the slope of \(f'\), so a rising slope bends the curve up.

You use it to describe a curve from the signs of its derivatives, or to read those signs from a graph.

This is one of five skills in topic 5.9 of Which & Why: AP Calculus.