Reading Shape from the Sign of the Derivative

You already know \(f'\)(c) is the slope of the tangent line at \(x = c\) — the number the limit definition produces. This lesson reads just its sign: where \(f' > 0\) the graph of \(f\) rises, where \(f' < 0\) it falls, and where \(f' = 0\) the tangent is horizontal. That's enough to sketch either graph from the other.

By the end you'll be able to read where \(f\) increases, decreases, and has horizontal tangents straight off the sign of \(f'\) — and sketch the graph of \(f'\) from the graph of \(f\) and vice versa.

Predict: where the top curve is falling, what sign will the bottom curve have? And what value must \(f'\) hit at the peak and at the valley of \(f\)? Drag the slider and watch the bottom curve cross zero exactly at the two horizontal tangents.

Top: \(f(x) = x^3 - 3x\). Bottom: its derivative \(f'(x) = 3x^2 - 3\), on the same x-axis. The dashed line marks one \(x\) value on both graphs: the tangent segment on \(f\) and the dot on \(f'\) share a color because they are the same number — the height of \(f'\) is the slope of \(f\). Hover or focus the amber markers and the dot for exact values.

x = 0.00 · slope f′(0.00) = −3.000 → f is decreasing ↘

f above, f′ below — green where f rises (f′ > 0), red where f falls (f′ < 0), amber at the horizontal tangents (f′ = 0)
f increasing (f′ > 0) f decreasing (f′ < 0) horizontal tangent (f′ = 0) graph of f′

The sign and magnitude of \(f'\) describe the shape of \(f\): increasing where \(f' > 0\), decreasing where \(f' < 0\), horizontal tangents where \(f' = 0\) — so each graph can be sketched from the other.

Visual

Scan the graph of \(f\) left to right and transcribe what you see: while \(f\) climbs, plot \(f'\) above its axis; while \(f\) falls, plot \(f'\) below; at each critical point — a peak, valley, or momentary plateau — \(f'\) touches zero. Magnitude carries over too: a steep stretch of \(f\) means \(f'\) is far from the axis; a nearly flat stretch means \(f'\) hugs it. Going the other way, wherever \(f'\) sits above the axis draw \(f\) uphill, wherever it sits below draw \(f\) downhill, and wherever \(f'\) crosses the axis with a genuine sign change, \(f\) turns around.

Intuitive

Think of \(f'\) as a speedometer with a direction needle. Positive reading: you're driving uphill. Negative: downhill. Zero: level road, for at least an instant. The bigger the reading, the steeper the road. A peak in your altitude (the graph of \(f\)) is the moment the needle sweeps from positive through zero to negative — the needle itself doesn't peak, it crosses zero. That's why in the interactive the dot on \(f'\) sits at exactly zero when the tangent up top goes flat.

Worked example

Let \(f(x) = x^2\), so \(f'(x) = 2x\). Read the sign of \(2x\): for \(x < 0\), \(f'(x) < 0\), so \(f\) is decreasing on the left. At \(x = 0\), \(f'(0) = 0\) — a horizontal tangent, the vertex. For \(x > 0\), \(f'(x) > 0\), so \(f\) is increasing on the right. The sign change of \(f'\) from − to + at \(x = 0\) confirms a minimum there — exactly the familiar upward parabola.

Your turn

Let \(g(x) = x^2 - 4x\), so \(g'(x) = 2x - 4\). The sign of \(2x - 4\): negative for \(x < 2\), so \(g\) is decreasing there; \(g'(2) = 0\), a horizontal tangent at \(x = 2\). Now finish it: for \(x > 2\), \(g'(x)\) is ____, so \(g\) is ____, and the sign change of \(g'\) at \(x = 2\) means \(g\) has a ____ there.

Reveal the answer

For \(x > 2\), \(2x - 4 > 0\), so \(g\) is increasing. \(g'\) changes sign from − to +, so \(g\) has a minimum at \(x = 2\) (it's the parabola \(x^2 - 4x\) with vertex at \(x = 2\)). Same three-step read as the worked example: sign left, zero, sign right.

More info — why a peak of f is a zero crossing of f′, not a peak of f′

The most common translation error is copying features of \(f\) onto \(f'\). But the graphs answer different questions: \(f\) says "how high am I?", \(f'\) says "how fast am I climbing?". Approaching a peak you're still climbing (\(f' > 0\)), just slower and slower — so \(f'\) is falling toward zero while \(f\) is still rising. Past the top you descend (\(f' < 0\)). Net effect: the peak of \(f\) is where \(f'\) crosses the axis from + to −. Set the slider to the peak at \(x = -1\) in the interactive above and watch the dot on \(f'\) sit at zero mid-crossing. And remember the flip side from the quiz: a zero of \(f'\) without a sign change (like \(x^3\) at 0) is a flat instant, not a turn — Paul's Online Notes in Dive deeper below works several of these sign-chart cases in full.

Check your understanding

Question 1 of 4

A function \(f\) has derivative \(f'(x) = 2x - 6\). What does this tell you about the shape of \(f\)?

Question 2 of 4

If \(f'(c) = 0\), must \(f\) have a maximum or minimum at \(c\)?

Question 3 of 4

At \(x = 4\) the tangent line to the graph of \(f\) has slope \(-2\). What is \(f\) doing at \(x = 4\)?

Question 4 of 4

The graph of \(f\) has a smooth peak (local maximum) at \(x = 2\). What does the graph of \(f'\) do at \(x = 2\)?

Recap

  • \(f' > 0\) on an interval → \(f\) is increasing there; \(f' < 0\) → decreasing.
  • \(f' = 0\) → horizontal tangent: a candidate peak, valley, or plateau (a critical point). It's a real max/min only if \(f'\) changes sign — \(x^3\) at 0 is the standard counterexample.
  • Magnitude matters too: large \(|f'|\) means a steep graph of \(f\); small \(|f'|\) means nearly flat.
  • Translate both ways: a peak/valley of \(f\) is a zero crossing of \(f'\); the height of \(f'\) at each \(x\) is the tangent slope of \(f\) there.

Dive deeper

Sources

  • Reading Shape from the Sign of the Derivative