Lessons 8 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
8
Investigating functions and sketching graphs with the derivative
Textbook, Part 1: pp. 42–49
GoalUse the derivative to find intervals of increase and decrease, stationary points, local extrema and greatest/least values on a segment, and to sketch a graph.
New words
stationary point · statsionar nuqtalocal maximum / minimum · lokal maksimum / minimumextremum · ekstremumgreatest / least value · eng katta / eng kichik qiymat
Explanation
If f′(x) > 0 on (a, b), then f increases there; if f′(x) < 0, it decreases. Points where the derivative is zero are called stationary points. If f′ changes sign from + to − when passing through a stationary point, it is a local maximum; from − to +, a local minimum; if the sign does not change there is no extremum (for example the stationary point x = 0 of f(x) = x³). To find the greatest and least values of a function continuous on a segment, compute its values at the stationary points inside the segment and at the endpoints and pick the largest and smallest. Order for sketching a graph: domain, stationary points, intervals of increase and decrease, extrema and their values, intersections with the axes, then the picture.
Worked examples
f(x) = x³ + 3x² − 9x − 5. f′(x) = 3x² + 6x − 9 = 3(x + 3)(x − 1). f′ > 0 on (−∞, −3) and (1, ∞): increasing; f′ < 0 on (−3, 1): decreasing. At x = −3 the sign goes from + to −: local maximum f(−3) = 22. At x = 1 it goes from − to +: local minimum f(1) = −10. The graph meets the Oy axis at (0, −5).
The greatest and least values of f(x) = 2x³ − 9x² + 12x on [0, 3]. f′ = 6x² − 18x + 12 = 6(x − 1)(x − 2); the stationary points 1 and 2 lie in the segment. f(0) = 0, f(1) = 5, f(2) = 4, f(3) = 9. The greatest value is 9 (at x = 3) and the least is 0 (at x = 0).
Class activity
“Reading the derivative graph”: the teacher draws the graph of f′(x) with its zeros and signs marked. Groups find the intervals where f increases and decreases and its local maxima and minima, sketch f, and compare with other groups.
Practice
1
Find the stationary points of f(x) = x³ − 12x.
f′ = 3x² − 12 = 0 ⇒ x = −2 and x = 2.
2
On which interval does f(x) = x² − 6x + 5 increase?
(3, ∞)
3
Find the greatest and least values of f(x) = x³ − 3x² on [−1, 3].
f′ = 3x² − 6x = 3x(x − 2) ⇒ x = 0, 2 (in the segment). f(−1) = −4, f(0) = 0, f(2) = −4, f(3) = 0. The greatest value is 0, the least is −4.
4
Is a stationary point always an extremum? Explain using f(x) = x³.
No. f′ = 3x² = 0 at x = 0, but f′ ≥ 0 everywhere and the sign does not change: the function keeps increasing and has no extremum.