Topic 5.9 key terms
- graph of f′
- The graph whose height at each x is the slope of f at that x. Its sign tells you whether f is increasing or decreasing; whether it is rising or falling tells you the concavity of f.
- second derivative f″
- The derivative of f′: the slope of the graph of f′. It tells you how the slopes of f are changing.
- relative maximum (from f′)
- If f is continuous at c, f′ changing from positive to negative at c gives f a relative maximum there (the first derivative test).
- relative minimum (from f′)
- If f is continuous at c, f′ changing from negative to positive at c gives f a relative minimum there.
- concave down
- The slopes of f are decreasing: f′ is decreasing. f″ < 0 throughout an interval is enough to guarantee it; an isolated zero of f″ need not change the concavity.
- concave up
- The slopes of f are increasing: f′ is increasing. f″ > 0 throughout an interval is enough to guarantee it (x⁴ is concave up even though f″(0) = 0).
- point of inflection
- A point where the graph of f changes concavity: f′ changes from increasing to decreasing or back (f″ changes sign). f″ may fail to exist there.
- critical point
- An input in the domain of f where f′ is zero or does not exist. It is only a candidate for a relative maximum or minimum; if f is continuous there, f′ changing sign settles it.
- inflection candidate
- An input where f″ is zero or does not exist. It is an inflection point only if f is continuous there and its graph changes concavity (f′ changes between increasing and decreasing).
- same derivative, shifted graph
- f and f + C have the same derivative, so f′ fixes the shape of f; one known value of f fixes its height.
Transcript
965 words · 3 parts
The three-pen machine
Let’s watch all three graphs come alive at once. This machine has a cart on a track and three pens, and as the cart moves, the paper rolls out and each pen draws one graph.
Let’s start with a run that follows our same friendly curve, just stretched out over eight seconds. The top pen draws the cart’s position, which we’ll call f, the middle one draws its velocity, f prime, and the bottom one draws its acceleration, f double prime. The velocity pen draws the slope of the position graph, and the acceleration pen does the same job for velocity.
At two seconds the cart stops and turns around, so the position graph has a flat tangent, and directly below it the velocity is exactly zero. At four seconds the velocity bottoms out, so its tangent is flat and the acceleration crosses zero, and up top that’s exactly where the position graph bends.
Now let’s make it harder. In this run the cart bounces on a spring, and somebody tore the labels off the printout and shuffled the three strips. Which strip is f, which one is f prime, and which is f double prime? Find a strip with a flat tangent and look for the strip that’s zero at that same moment, then check the signs, because a matching flat spot is only a candidate.
Now you get to drive. Your stick sets the velocity, f prime, and the dashed hill is the position graph we want the blue pen to draw. Push the stick to set f prime and make the blue pen trace the hill. Once you’ve matched it, the machine hands you a blank page, and then it’s your turn to design a trip that turns around twice.
On the board, that hill is f of t equals t minus t squared over eight, and its derivative is the straight line one minus t over four. It’s positive, then zero at four, then negative, which is the stick move that draws the hill.
On the exam
This is exactly how the AP exam asks it. They hand you the graph of f prime, here three straight line segments, and they never show you f at all. Part a asks where f has a relative maximum, and wants a justification. We’re looking for places where f prime crosses from positive to negative, and that happens twice, at x equals one and at x equals five.
Let’s write the sentence that earns the point, and notice that it names f prime and says which way the sign changes. f has a relative maximum at x equals one and at x equals five, because f prime changes from positive to negative at each of them. The same reasoning at x equals three, where f prime goes from negative to positive, gives a relative minimum. Let’s do part b together. On what open interval is the graph of f concave up?
Part c wants the inflection points. The graph of f changes concavity wherever f prime switches between going down and going up, and that happens at the two corners, x equals two and x equals four.
Notice that f double prime doesn’t even exist at those sharp corners, and that’s fine. It jumps from negative two to two at x equals two, and what makes an inflection point is that the concavity really changes.
For part d they tell us f of zero equals one. Use the tangent line at x equals zero to estimate f of one half. Is two too big or too small? On zero to two f prime is decreasing, so f is concave down, and a concave down curve sits below its tangent lines, which makes two an overestimate.
Let’s do one last question on this graph, and this time you’ll need both readings at once. On what open interval is f decreasing and concave up at the same time?
Sometimes the exam doesn’t give you a graph at all, just a table like this one, with values of f prime and f double prime at a few points, and that’s where last lesson’s second derivative test comes in. At x equals two, f prime is zero, so the tangent is flat, and f double prime is negative, so f is concave down right there. That makes the flat spot the top of a hill, a relative maximum. The same reasoning at four, where f double prime is positive, gives a relative minimum.
Sharp
Let’s go back to the mountain, because the computer has one more slice for you, and this time you’re the one who gets to fix the star. Here are the slopes, and this time you get to draw the light yourself. Read them the way we’ve been reading them all lesson, and give your light a crest, a trough and a bend in the right places. Drag the handles to shape the blue light until its slopes match the orange graph, and keep an eye on the star while you do it, because the closer you get, the sharper it gets.
Let’s put the whole lesson in a few lines. The height of f prime is the slope of f, the sign of f prime tells you whether f is climbing or falling, and whether f prime is rising or falling tells you which way f bends. Where f prime changes sign, f has a peak or a valley, an inflection point sits where f prime switches between increasing and decreasing, and one point on f pins down the height.
At the start I promised you’d be able to look at a graph of slopes and draw the shape it came from. You just did it, the same way that telescope does it a thousand times every second.