Topic 5.9 Notes – Connecting a Function, Its First Derivative, and Its Second Derivative

How f, f′, and f'' Are Connected

Everything comes down to sign (positive/negative) and increasing or decreasing behavior.

Core relationships

• If $f′(x) > 0$ → $f$ is increasing
• If $f′(x) < 0$ → $f$ is decreasing
• If $f′(x) = 0$ → critical point of $f$ (possible max/min)

• If $f''(x) > 0$ → $f$ is concave up and $f′$ is increasing
• If $f''(x) < 0$ → $f$ is concave down and $f′$ is decreasing
• If $f''(x) = 0$ → possible inflection point (must check sign change)

Think of it like layers:

• $f′$ controls up/down movement
• $f''$ controls bending

The three graphs below show the same function and its first two derivatives lined up side by side.

Aligned graphs of $f(x)$, $f′(x)$, and $f''(x)$

Notice how the x-values of the local maxima and minima of $f$ match the zeros of $f′$. Then look at where $f′$ has its own peaks and valleys. Those line up with zeros of $f''$. When you study graphs like this, you are matching features across layers.

Increasing, Decreasing, and Relative Extrema

Reading increasing/decreasing from $f′$

If you’re given the graph of $f′$:

• Above the x-axis → $f′ > 0$ → $f$ increasing
• Below the x-axis → $f′ < 0$ → $f$ decreasing

That’s it. No slope estimation needed.

Relative maxima and minima

All relative extrema of $f$ occur where:

• $f′ = 0$ or undefined

But zero alone isn’t enough. You need a sign change.

• $f′$ changes positive → negative → relative maximum
• $f′$ changes negative → positive → relative minimum

Here’s a clean example. This is a graph of $f'(x) = x^3 - 4x$, so we are using only the derivative to predict what $f$ is doing.

From this graph alone, look at how $f′$ changes sign at each zero:

• At $x=-2$, $f′$ changes negative → positive, so $f$ has a relative minimum.
• At $x=0$, $f′$ changes positive → negative, so $f$ has a relative maximum.
• At $x=2$, $f′$ changes negative → positive, so $f$ has another relative minimum.

On FRQs, the justification must mention the sign change. Just writing “because $f′=0$” will not earn the point.

Concavity and Points of Inflection

Concavity from $f''$

If you’re given the graph of $f''$:

• Above x-axis → $f'' > 0$ → concave up
• Below x-axis → $f'' < 0$ → concave down

You can also read concavity from $f′$:

• If $f′$ is increasing → $f'' > 0$ → concave up
• If $f′$ is decreasing → $f'' < 0$ → concave down

Points of inflection

A point of inflection occurs where:

• $f$ changes concavity
• which means $f''$ changes sign

Major connection:

• Inflection points of $f$ are relative extrema of $f′$
• They are also x-intercepts of $f''$ with sign change

If $f''$ just touches zero and stays the same sign, there is no inflection point. This is a common trap on multiple choice.

The Full Relationship Map

Here’s the whole chain in one place:

• Relative extrema of $f$
  ↔ x-intercepts of $f′$ (with sign change)
• Points of inflection of $f$
  ↔ relative extrema of $f′$
  ↔ x-intercepts of $f''$ (with sign change)
• Increasing/decreasing of $f$
  ↔ sign of $f′$
• Concavity of $f$
  ↔ sign of $f''$
  ↔ increasing/decreasing of $f′$

When you’re given only one graph on a test, mentally build the others.

If they give you $f′$, you should automatically be thinking about:

• where it crosses zero,
• where it’s positive or negative,
• where it’s increasing or decreasing.

That alone tells you almost everything about $f$.