Curve Analyser

Find where a function has maxima and minima, where it increases and decreases, its points of inflection, and its absolute maximum and minimum on an interval.

f(x) =	x^3 - 6x^2 + 9x + 1
From x =	-1
To x =	5

f(x) =

Use x, + − * / ^, brackets, sin, cos, ln, sqrt, exp, π and e.

From x =

To x =

The closed interval for the absolute maximum and minimum, and where to look for turning points.

Turning points of x^3 − 6x^2 + 9x + 1

max at x = 1, min at x = 3

f is increasing on (−1, 1); f is decreasing on (1, 3); f is increasing on (3, 5). On [−1, 5] the absolute maximum is 21 at x = 5 and the absolute minimum is −15 at x = −1.

Exact fraction. Checked numerically.

Method: f′ = 0 and the second-derivative test
Worked out with: AllCalcHub symbolic derivative, numerical roots
Check: Every step that can be checked was checked; f′ evaluated at every critical point and found to be 0

x = 1: local maximum, f = 5
x = 3: local minimum, f = 1
Points of inflection: x = 2
Absolute maximum on the interval: 21 at x = 5
Absolute minimum on the interval: −15 at x = −1

y = x^3 − 6x^2 + 9x + 1

Maxat (1, 5)Point at the chart, or Tab to it and use ← →

Max: (1, 5)Min: (3, 1)Inflection: (2, 3)

Where f increases and decreases

Interval	Sign of f′	f is
(−1, 1)	+	increasing
(1, 3)	−	decreasing
(3, 5)	+	increasing

About the Curve Analyser

At a turning point the slope is zero, so start by solving f′(x) = 0. The second-derivative test then classifies each point: f″ < 0 means a local maximum, f″ > 0 a local minimum. If f″ is 0, look at the sign of f′ either side instead.

On a closed interval, the absolute maximum and minimum are among the critical points and the two ends, so compare f at all of them.

The formula

f′(c) = 0; f″(c) < 0 ⇒ maximum, f″(c) > 0 ⇒ minimum

Worked example

f(x) = x³ − 6x² + 9x + 1

f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3)
f″(1) = −6 < 0: maximum f(1) = 5
f″(3) = 6 > 0: minimum f(3) = 1

Questions people ask

Why give an interval?

The absolute maximum and minimum depend on where you look, and turning points are searched for in that interval.

What's the difference between a critical point and a turning point?

A critical point is where f′(x) = 0 or f′ doesn't exist. It's a turning point only if f′ changes sign there: x³ has a critical point at 0 but keeps increasing through it.