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What the first derivative's sign says, what the second's says, and reading both at once.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the intervals where a function increases and decreases from the sign of its derivative, build a sign chart split at every point where the derivative vanishes or fails to exist, find inflection points by checking that the second derivative changes sign rather than merely vanishes, read the four combinations of rise and bend as descriptions of real behaviour, and translate a verbal description of a curve back into statements about $f'$ and $f''$.
The first derivative gives the direction. On an interval where $f' > 0$, $f$ is increasing; where $f' < 0$, decreasing. This is not a definition and not obvious — it is the mean value theorem: $f(q) - f(p) = f'(c)(q-p)$, and if $f'$ is positive throughout then the right side is positive whenever $q > p$.
The second derivative gives the bend. Where $f'' > 0$ the graph is concave up (it holds water, and lies above its tangents); where $f'' < 0$, concave down. An inflection point is where the concavity changes.
The two are independent, so all four combinations happen:
| $f'' > 0$ (bends up) | $f'' < 0$ (bends down) | |
|---|---|---|
| $f' > 0$ (rising) | rising, ever faster | rising, levelling off |
| $f' < 0$ (falling) | falling, levelling off | falling, ever faster |
"Falling and concave up" is not a contradiction; it is a ball rolling into the bottom of a bowl.
Sign charts. To find where $f'$ is positive, locate every point where $f'$ is zero or undefined, split the line at all of them, and test one point in each piece. Both kinds of point split the line, and forgetting the second kind merges two intervals that genuinely differ.
Vanishing is not changing. $f''(c) = 0$ does not make $c$ an inflection: $x^4$ has $f''(0) = 0$ and is concave up on both sides. The sign must actually change.
Another way: picture
Two sign charts drawn one above the other along the same axis — $f'$ on top, $f''$ below. Read a column and you have the shape there: the top row says up or down, the bottom row says which way it curves. Every feature worth marking on a sketch is a place where one of the two rows changes.
Another way: steps
$f''$ measures how fast $f'$ is changing, and reading it that way makes the vocabulary unnecessary.
If $s(t)$ is position, $s'$ is velocity and $s''$ is acceleration. Concave up means the velocity is increasing — which, for a falling object, means it is speeding up, and for a rising one means it is speeding up too. The sign of $s''$ says nothing about which way you are going.
The distinction is worth having in ordinary language, because it is routinely muddled. "Inflation is falling" is a statement about the second derivative of prices: prices are still rising, just less quickly. Someone who hears it as "prices are falling" has read $f'' < 0$ as $f' < 0$.
The same confusion, in an epidemic: the inflection point of the cumulative case curve is the day the daily number peaks. The total is still climbing, and will climb for a long time yet. The inflection is good news about the rate and not yet news about the total.
Reading $f'' = 0$ as an inflection. The sign has to change. $x^4$ at $0$ is the counterexample, and it is the same shape of error as reading $f' = 0$ as an extremum.
Leaving undefined points out of a sign chart. $f'(x) = 1/x^2$ is positive either side of $0$ and the point still splits the domain; $f'(x) = (x-1)/x$ changes sign at $0$ with no zero there.
Testing the sign of $f$ instead of $f'$. Increasing is about the derivative. A function can be negative and increasing throughout.
Assuming concave up means increasing. Independent questions, four combinations.
Treating a turning point as necessarily where $f' = 0$. A corner turns too, and $f'$ does not vanish there. $|x|$ has a minimum at a point where the derivative does not exist.
Sketching from the formula rather than from the charts. The charts are the evidence; the sketch is a summary of them.
The words fight the meaning. "Concave up" describes the bend, and a graph can bend upwards while heading firmly downwards — every falling object approaching terminal velocity does exactly that.
The reliable translation is into rates. $f' $ is the rate; $f''$ is the rate at which the rate changes. "Concave up" is "the rate is increasing", and a rate can increase from $-10$ to $-3$ without ever becoming positive.
This is worth fixing here because the integral unit inherits it. The fundamental theorem says an accumulation function has the integrand as its derivative, so the integrand's sign says whether the accumulation rises and the integrand's slope says how it bends — two levels of the same confusion, and the second one is harder to see through.
$f(x) = x^3 - 3x^2$. $f'(x) = 3x^2 - 6x = 3x(x - 2)$, zero at $x = 0$ and $x = 2$.
Two candidates.
Sign chart for $f'$: positive on $x<0$, negative on $0<x<2$, positive on $x>2$. So a maximum at $0$ and a minimum at $2$.
Test one point in each piece.
$f''(x) = 6x - 6$, zero at $x = 1$ and changing sign there: concave down on $x<1$, up on $x>1$, inflection at $(1, -2)$. The inflection sits between the two turning points, as it must — the bend has to reverse to get from a maximum to a minimum.
$f(x) = x^4$. $f''(x) = 12x^2$, which is zero at $x = 0$.
The candidate.
But $12x^2 \ge 0$ everywhere, so the concavity is up on both sides and never changes.
Check the sign, not the value.
No inflection at $0$. It is a minimum, and a notably flat one — the graph hugs the axis near the origin, which is what a vanishing second derivative at a minimum looks like.
$f'(x) = 3x^2 - 12 = 3(x-2)(x+2)$, zero at $\pm 2$.
Factor to read the sign.
Test points: at $x = -3$, $f' = 15 > 0$; at $0$, $f' = -12 < 0$; at $3$, $f' = 15 > 0$.
One point per piece.
Increasing on $x < -2$ and on $x > 2$; decreasing between. Two intervals, not one.
Match each derivative sign observation to the graph feature it establishes.
| Increasing | Decreasing | Concave up | Inflection point at $c$ | |
|---|---|---|---|---|
| $f'>0$ | ||||
| $f'<0$ | ||||
| $f''>0$ | ||||
| $f''$ changes sign at $c$ |
Match each derivative-sign pattern to the graph behavior it proves.
| Increasing and concave down | Decreasing and concave up | A local maximum at $c$ | An inflection point at $d$ | |
|---|---|---|---|---|
| $f'>0$ and $f''<0$ | ||||
| $f'<0$ and $f''>0$ | ||||
| $f'$ changes from positive to negative at $c$ | ||||
| $f''$ changes from negative to positive at $d$ |
On which interval is $f(x) = 5x^2 - 20x$ increasing? Give it as an interval.
This task has no paper form; do it on a device.
Where is the inflection point of $f(x) = 2x^3 - 4x^2 + x$?
Answer:
At a point, $f' < 0$ and $f'' > 0$. What is the graph doing?
| Rising and getting steeper | Rising and flattening | Falling and flattening | Falling and getting steeper | |
|---|---|---|---|---|
| $f'>0$, $f''>0$ | ||||
| $f'>0$, $f''<0$ | ||||
| $f'<0$, $f''>0$ | ||||
| $f'<0$, $f''<0$ |
A population rises, ever more slowly, towards a ceiling it never reaches. How many inflection points does its graph have, if it starts by rising ever faster?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
To build a sign chart for $f'(x) = \dfrac{x - 8}{x}$, which points split the line?
| May change the derivative sign | Must also split a sign chart | Has a constant derivative sign if no split point lies inside | Determines the sign on that interval | |
|---|---|---|---|---|
| A point where $f'=0$ | ||||
| A point where $f'$ is undefined | ||||
| An interval between consecutive split points | ||||
| One interior test point |
You can build sign charts for $f'$ and $f''$ and read a graph's shape from them. Say in your own words why a falling graph can be concave up. Next: applied optimisation, where the function has to be built before it can be maximised.
8. Your turn: where is $f(x) = x^3 - 12x$ increasing?, step 3