Increasing, Decreasing and the First Derivative Test

Main idea

Where $f'$ is positive the graph goes up, and where it's negative the graph goes down. So at a critical point, just look at the slope on either side: up then down is a peak, down then up is a valley, and no change means neither.

f' > 0 \text{ on an interval} \implies f \text{ increasing there}, \qquad f' < 0 \implies f \text{ decreasing}

First derivative test

f' \text{ changes } + \to - \text{ at } c \implies \text{relative max}; \qquad - \to + \implies \text{relative min}

Explanation

The first derivative test is simply a way to tell whether a critical point is a peak, a valley or neither, by checking which way the graph is heading on each side of it. All we're really saying is: if you walk uphill and then downhill, you went over a peak. The formulas are just that sentence in symbols: $f' > 0$ (going uphill, so $f$ is increasing), $f' < 0$ (going downhill, so decreasing), and " $+ \to -$ at $c$ " (the slope switches from uphill to downhill as you pass $c$ ). Nothing more to it than that.

Two lessons ago we found critical points, and we saw that some of them, like $0$ for $x^3$ , aren't peaks or valleys at all. So how do we tell them apart? The sign of $f'$ already tells you whether the graph is going up or down. Check it just before and just after the critical point, and you know what the graph did there.

Sign charts. The critical points split the number line into intervals. On each interval, $f'$ keeps one sign the whole way, because the only places it can switch from positive to negative are where it's zero or undefined, and those are exactly the critical points. So you only need to test one point in each interval.

Then read off:

$f'$ goes from $+$ to $-$ : the function climbs then falls, a relative max.
$f'$ goes from $-$ to $+$ : it falls then climbs, a relative min.
No sign change: neither, like $x^3$ at 0, which climbs, pauses, and keeps climbing.

See it

This is a drug's concentration in the blood, C(t) = 20te^(−0.5t), drawn with its derivative C′(t). As you can see, C′ is positive for the first 2 hours, so the level climbs, and then it turns negative, so the level falls. That switch from + to − at t = 2 marks the peak, about 14.7 mg/L.

Worked examples

Example 1

Find where $f(x) = 2x^3 + 3x^2 - 12x$ increases and decreases, and classify its critical points.

1
$f'(x) = 6x^2 + 6x - 12 = 6(x + 2)(x - 1)$ , so the critical points are $-2$ and $1$ . They split the line into three intervals.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Where is $f(x) = x^2 - 6x$ decreasing?

Problem 2+5 XP

Where does $f(x) = x^3 - 12x$ have a relative maximum?

Problem 3+5 XP

Where does $f(x) = x^3 - 12x$ have a relative minimum?

Problem 4+5 XP

For $f(x) = x^4 - 4x^3$ , what kind of point is $x = 0$ ?

Problem 5+5 XP

On which interval is $f(x) = xe^{-x}$ increasing?

Problem 6+5 XP

A stock's price over six months follows $P(t) = t^3 - 9t^2 + 24t + 50$ dollars, $t$ months in. When did it reach a relative (local) maximum?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

For which values of $k$ is $f(x) = x^3 + kx^2 + 3x$ increasing on the whole real line? The answer is an interval $-a \le k \le a$ ; find $a$ .

Where it's used

A drug's concentration curve rises while absorption outpaces elimination and falls afterward; the peak (where $C'$ changes sign) sets the dose's strongest effect.

IV Drip ConcentrationDiff Eq · Medicine

Finding Absolute Extrema Concavity and the Second Derivative Test