Draw 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.
A local extremum of $f$ occurs at a ____ of $f'$ where it changes sign.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A sign-changing zero of $f'$ marks an extremum of $f$.
Which features come from $f''$ (the second derivative)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Concavity + inflection from $f''$; increase/decrease + extrema from $f'$.
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 $f'$.
Where the graph of $f$ has a horizontal tangent, the graph of $f'$...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Horizontal tangent ⇒ slope $0$ ⇒ $f'=0$.
Where $f$ is increasing, its derivative $f'$ is positive.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Rising ⇔ $f'>0$.
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.
The finished shape
y = ax³ + cx
Peak at $(-1,2)$, inflection at $(0,0)$, valley at $(1,-2)$ — exactly what the $f'$ and $f''$ charts predict.
At a point where $f$ is high but flat, the value of $f'$ is...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Read the slope, not the height: flat ⇒ $f'=0$.
A local maximum should occur where $f$ turns from increasing to decreasing and the curve is concave...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A peak is concave down (a cap).
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.