Before Quiz 2
What does the graph of f′ say about the shape of f?
Read a curve from its derivative: the sign of f′ says up or down, a change of sign makes a peak or a valley, and the rise and fall of f′ set the bend.
Builds on the chain rule and logarithms and limits.
The lesson as text
1. Up where f′ is above
Above is a curve ; below it, its derivative , on the same . The tangent at the probe has slope , which the lower panel draws as a height. Where is above the axis, rises; where it is below, falls.
Drag the probe to the bottom of the valley on the right of .
There crosses the axis from below to above, at : stops falling and starts rising, so it has a minimum. Where crosses from above to below, at , has a maximum.
The figure: The curve f(x) = x⁴/4 − 3x²/2 + 0.50x. At the probe, x = 1.64, f is falling and its tangent has slope −0.01. Its derivative f′(x) = x³ − 3x + 0.50, which at the probe has height −0.01. Its lowest point on the right is at x = 1, height −1.50.
2. The bend is the slope of f′
Between the peak and the valley, changes from bending down to bending up. The bend is how its slopes change: while falls, the slopes of steepen downward; while rises, they turn back up.
Question: Where does switch from bending down to bending up?
Where is lowest, at . To its left falls, so and bends down; to its right rises and bends up. Here , zero at . Between the peak and the valley it is also where falls fastest: .
The figure: The curve f(x) = x⁴/4 − 3x²/2 + 0.50x. At the probe, x = 1.00, f is falling and its tangent has slope −1.50. Its derivative f′(x) = x³ − 3x + 0.50, which at the probe has height −1.50. Its lowest point on the right is at x = 1, height −1.50.
3. Lift f′
Now lift by dragging its lowest point. Adding to adds to , which tilts it. Watch the peak and the valley on the right of as the dip of rises towards the axis.
Drag the dip of up until it just touches the axis.
At the dip touches the axis at : . The peak and the valley closed in on each other and met there, and the tangent at is flat.
The figure: The curve f(x) = x⁴/4 − 3x²/2 + 2.00x. At the probe, x = 1.00, f is flat and its tangent has slope 0.00. Its derivative f′(x) = x³ − 3x + 2.00, which at the probe has height 0.00. Its lowest point on the right is at x = 1, height 0.00.
4. Flat is not a turn
Here , so is a critical point. A common slip stops there and reports a maximum or a minimum. In amber: the cap a maximum at would make, and the it would need, crossing the axis from above to below.
Question: What does have at ?
Neither. is positive just left of and just right of it, so rises into the flat point and keeps rising. A zero of only marks a candidate; a change of sign makes a turn. At , does change sign, from negative to positive: a minimum, . The second derivative test is silent at , as .
The figure: The curve f(x) = x⁴/4 − 3x²/2 + 2.00x. At the probe, x = 1.00, f is flat and its tangent has slope 0.00. In amber, the cap that a maximum at x = 1 would make; the curve itself passes through (1, 0.75) rising on both sides. Its derivative f′(x) = x³ − 3x + 2.00, which at the probe has height 0.00. Its lowest point on the right is at x = 1, height 0.00. In amber, the f′ a maximum would need: crossing the axis at x = 1 from above to below.