The first derivative tells us if a function is increasing or decreasing. The second derivative tells us how the slope itself is changing — whether the function is bending upward (concave up) or downward (concave down). This helps us understand the shape of a graph and provides another way to classify local extrema.
🎯 In this section you will learn
- The meaning of concavity (concave up vs concave down)
- How to find intervals of concavity using $f''(x)$
- How to find inflection points (where concavity changes)
- The Second Derivative Test for local maxima and minima
📌 Concavity
$$ \text{If } f''(x) > 0 \text{ on an interval, then } f \text{ is } \textbf{concave up} \text{ (cup-shaped)} $$
$$ \text{If } f''(x) < 0 \text{ on an interval, then } f \text{ is } \textbf{concave down} \text{ (cap-shaped)} $$
💡 Intuition
- Concave up: The graph lies above its tangent lines. Slope is increasing.
- Concave down: The graph lies below its tangent lines. Slope is decreasing.
Find where $f(x) = x^3 - 3x$ is concave up and concave down.
①
Find $f''(x)$
$f'(x) = 3x^2 - 3$, $f''(x) = 6x$
②
Find where $f''(x)=0$
$6x = 0$ ⇒ $x = 0$ (possible inflection point)
③
Test intervals
On $(-\infty, 0)$: $f''(x) < 0$ → concave down
On $(0, \infty)$: $f''(x) > 0$ → concave up
📌 Inflection Points
An inflection point is where the concavity changes (from up to down or down to up). At an inflection point, $f''(x) = 0$ or $f''(x)$ is undefined, and the sign of $f''$ changes.
Find the inflection point of $f(x) = x^3 - 3x$.
- From Example 1, $f''(x) = 6x$, zero at $x=0$
- Concavity changes from down to up at $x=0$
- Inflection point: $(0, f(0)) = (0, 0)$
📌 Second Derivative Test
📌 Second Derivative Test
Suppose $c$ is a critical point of $f$ (so $f'(c)=0$).
- If $f''(c) > 0$, then $f$ has a local minimum at $c$.
- If $f''(c) < 0$, then $f$ has a local maximum at $c$.
- If $f''(c) = 0$, the test is inconclusive (could be max, min, or neither).
Use the Second Derivative Test to classify the critical points of $f(x) = x^3 - 3x$.
①
Critical points (from before)
$x = -1$ and $x = 1$
②
Evaluate $f''(x)=6x$
$f''(-1) = -6 < 0$ → local maximum at $x = -1$
$f''(1) = 6 > 0$ → local minimum at $x = 1$
Test $f(x) = x^4$ using the Second Derivative Test.
- $f'(x) = 4x^3$, critical point at $x=0$
- $f''(x) = 12x^2$, $f''(0) = 0$ — inconclusive!
- Actually $f(x)=x^4$ has a local minimum at $x=0$, but the second derivative test didn't detect it.
💡 Note
When $f''(c)=0$, you must use the First Derivative Test or examine higher-order derivatives.
Analyze $f(x) = x^4 - 4x^3$: find intervals of increase/decrease, concavity, and classify extrema.
①
Find derivatives
$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$
$f''(x) = 12x^2 - 24x = 12x(x-2)$
②
Critical points
$f'(x)=0$ ⇒ $x=0$, $x=3$
③
Increasing/Decreasing
Test: $f'(-1)=4(-1)^2(-4)=-16<0$ (decreasing)
$f'(1)=4(1)^2(-2)=-8<0$ (decreasing)
$f'(4)=4(16)(1)=64>0$ (increasing)
Decreasing on $(-\infty,3)$, increasing on $(3,\infty)$
④
Concavity
$f''(x)=0$ ⇒ $x=0$, $x=2$
$f''(-1)=12(-1)(-3)=36>0$ (concave up)
$f''(1)=12(1)(-1)=-12<0$ (concave down)
$f''(3)=12(3)(1)=36>0$ (concave up)
Concave up on $(-\infty,0)$ and $(2,\infty)$; concave down on $(0,2)$
⑤
Classify extrema
At $x=3$: $f'(x)$ changes from − to + ⇒ local minimum (also $f''(3)>0$)
At $x=0$: $f'$ does not change sign ⇒ not an extremum (inflection point)
⚠️ Important Notes
- $f''(x) > 0$ means the slope is increasing — the graph bends upward.
- $f''(x) < 0$ means the slope is decreasing — the graph bends downward.
- Inflection points require a change in concavity, not just $f''(x)=0$.
- The Second Derivative Test is faster but not always conclusive (when $f''(c)=0$, use First Derivative Test).
🔍 Key Takeaways
- $f''(x) > 0$ ⇒ concave up; $f''(x) < 0$ ⇒ concave down.
- Inflection points occur where concavity changes.
- Second Derivative Test: $f''(c) > 0$ ⇒ local min; $f''(c) < 0$ ⇒ local max.
- If $f''(c)=0$, the test is inconclusive — use the First Derivative Test instead.
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