Menggambar Grafik Fungsi dan Turunannya
DiperkenalkanDraw the whole shape from its derivatives
- The derivatives are a recipe for a graph: $f'$ gives the slopes, $f''$ gives the bends.
- Combine the two sign charts and you can sketch a curve you were never handed.
- Extrema come from $f'$; concavity and inflection come from $f''$.
- This is where all of Unit 5 comes together into one picture.
The information you assemble
- From $f'$: intervals of increase/decrease and the extrema (where $f'$ changes sign).
- From $f''$: intervals of concavity and the points of inflection (where $f''$ changes sign).
- Plus a few plotted points and any intercepts to anchor the height.
- Layer these together and the shape is forced.

Ekstrema lokal dari $f$ terjadi pada ____ dari $f'$ di mana ia berubah tanda.
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Nol yang berubah tanda dari $f'$ menandai ekstrema dari $f$.
Fitur mana yang berasal dari $f''$ (turunan kedua)?
Kesulitan 2/5 · 140 koin + bonus kecil pada penyelesaian pertama hari ini
Kekurvature + belok dari $f''$; peningkatan/pengurangan + ekstrema dari $f'$.
Sketching $f'$ from $f$, and back
- Given the graph of $f$: where $f$ has a horizontal tangent, $f'=0$ (a zero of $f'$).
- Where $f$ rises, $f'$ is positive; where $f$ falls, $f'$ is negative.
- Where $f$ is steepest, $f'$ is at an extreme; at an inflection of $f$, $f'$ has a max or min.
- Reverse the reading to sketch $f$ from a given $f'$.
Di mana grafik $f$ memiliki tangen horizontal, grafik $f'$...
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Tangen horizontal ⇒ kemiringan $0$ ⇒ $f'=0$.
Di mana $f$ meningkat, turunannya $f'$ positif.
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Meningkat ⇔ $f'>0$.
Put the marks on the sketch
- Mark each extremum (peak/valley), each inflection point (bend flip), and label the intervals.
- Match increasing+concave-up (curving up steeply) vs. increasing+concave-down (leveling off), and so on — four combinations.
- Check consistency: a local max should sit where increase turns to decrease and the curve is concave down.
- A clean sketch shows all four features agreeing.
Bentuk akhirnya
y = ax³ + cx
Puncak di $(-1,2)$, belok di $(0,0)$, lembah di $(1,-2)$ — persis apa yang diprediksi oleh grafik $f'$ dan $f''$.
Di titik di mana $f$ tinggi tetapi datar, nilai $f'$ adalah...
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Baca kemiringan, bukan ketinggian: datar ⇒ $f'=0$.
Maksimum lokal seharusnya terjadi di mana $f$ berubah dari meningkat menjadi menurun dan kurva cembung...
Kesulitan 1/5 · 100 koin + bonus kecil pada penyelesaian pertama hari ini
Puncak cembung ke bawah (topi).
Don't read $f'$ off the height of $f$ — read it off the slope. Where $f$ is high but flat, $f'=0$, not "large." And a zero of $f'$ marks a possible extremum of $f$; a zero of $f''$ marks a possible inflection of $f$ — keep the two levels straight.
Sketch clues for $f(x)=x^3-3x$.
- $f'=3(x-1)(x+1)$: increasing on $(-\infty,-1)$ and $(1,\infty)$, decreasing on $(-1,1)$; max at $(-1,2)$, min at $(1,-2)$.
- $f''=6x$: concave down for $x<0$, concave up for $x>0$; inflection at $(0,0)$.
- The sketch: rises to a peak at $(-1,2)$, falls through $(0,0)$, dips to $(1,-2)$, then rises.
To sketch a graph, combine what $f'$ says (increase/decrease and extrema) with what $f''$ says (concavity and inflection points), plus a few anchor points. Read $f'$ from the slope of $f$ (not its height); a zero of $f'$ is a possible extremum, a zero of $f''$ a possible inflection.