Concavity for MATH 139

Exam Relevance for MATH 139

Likelihood of appearing: High

Use second derivative: f double prime > 0 means concave up, < 0 means concave down. Find inflection points.

Lesson

What is Concavity?

Concavity describes the "curvature" of a function — whether it curves upward like a bowl or downward like a hill.

The Big Idea: The second derivative tells you which way a function is curving.


Concave Up vs Concave Down

Concave Up (Smiley Face 😊)

A function is concave up when it curves upward — like a bowl that holds water.

  • The slope is increasing (getting steeper upward or less steep downward)
  • $f''(x) > 0$
  • Tangent lines lie below the curve

Notice how the parabola $f(x) = x^2$ opens upward. Any tangent line you draw will be below the curve (except at the point of tangency).

Concave Down (Frowny Face â˜šī¸)

A function is concave down when it curves downward — like an upside-down bowl.

  • The slope is decreasing (getting less steep upward or steeper downward)
  • $f''(x) < 0$
  • Tangent lines lie above the curve

The parabola $f(x) = -x^2$ opens downward. Tangent lines lie above the curve.


The Second Derivative Test for Concavity

If... Then the function is...
$f''(x) > 0$ Concave UP on that interval
$f''(x) < 0$ Concave DOWN on that interval

Why Does This Work?

  • $f''(x) = $ rate of change of $f'(x) = $ rate of change of the slope
  • If $f''(x) > 0$, the slope is increasing → concave up
  • If $f''(x) < 0$, the slope is decreasing → concave down

Inflection Points

An inflection point is where concavity changes — the function switches from concave up to concave down (or vice versa).

How to Find Inflection Points

  1. Find $f''(x)$
  2. Set $f''(x) = 0$ and solve (also check where $f''(x)$ is undefined)
  3. Test points on either side to confirm concavity actually changes
  4. Find the $y$-coordinate: plug the $x$-value back into $f(x)$

âš ī¸ Important: $f''(x) = 0$ is necessary but NOT sufficient for an inflection point. You must verify the concavity actually changes!


Example 1: Finding Concavity Intervals

Find the intervals where $f(x) = x^3 - 6x^2 + 9x + 1$ is concave up and concave down.

Step 1: Find the second derivative

$f(x) = x^3 - 6x^2 + 9x + 1$

$f'(x) = 3x^2 - 12x + 9$

$f''(x) = 6x - 12$

Step 2: Find where $f''(x) = 0$

$6x - 12 = 0$

$x = 2$

Step 3: Test intervals

Interval Test point $f''(x) = 6x - 12$ Concavity
$(-\infty, 2)$ $x = 0$ $6(0) - 12 = -12 < 0$ Down
$(2, \infty)$ $x = 3$ $6(3) - 12 = 6 > 0$ Up

$$\boxed{\text{Concave up: } (2, \infty) \quad \text{Concave down: } (-\infty, 2)}$$


Example 2: Finding Inflection Points

Find the inflection point(s) of $f(x) = x^4 - 4x^3 + 6$.

Step 1: Find the second derivative

$f'(x) = 4x^3 - 12x^2$

$f''(x) = 12x^2 - 24x = 12x(x - 2)$

Step 2: Find where $f''(x) = 0$

$12x(x - 2) = 0$

$x = 0$ or $x = 2$

Step 3: Test concavity around each point

Interval Test point $f''(x) = 12x(x-2)$ Concavity
$(-\infty, 0)$ $x = -1$ $12(-1)(-3) = 36 > 0$ Up
$(0, 2)$ $x = 1$ $12(1)(-1) = -12 < 0$ Down
$(2, \infty)$ $x = 3$ $12(3)(1) = 36 > 0$ Up

Concavity changes at both $x = 0$ and $x = 2$ ✓

Step 4: Find the $y$-coordinates

$f(0) = 0 - 0 + 6 = 6$

$f(2) = 16 - 32 + 6 = -10$

$$\boxed{\text{Inflection points: } (0, 6) \text{ and } (2, -10)}$$


Example 3: When $f''(x) = 0$ is NOT an Inflection Point

Show that $f(x) = x^4$ has no inflection point at $x = 0$.

Step 1: Find the second derivative

$f'(x) = 4x^3$

$f''(x) = 12x^2$

Step 2: Find where $f''(x) = 0$

$12x^2 = 0 \implies x = 0$

Step 3: Test concavity

Interval Test point $f''(x) = 12x^2$ Concavity
$(-\infty, 0)$ $x = -1$ $12(1) = 12 > 0$ Up
$(0, \infty)$ $x = 1$ $12(1) = 12 > 0$ Up

Concavity does NOT change! Both sides are concave up.

$$\boxed{\text{No inflection point at } x = 0}$$


Example 4: Concavity with Rational Functions

Find the intervals of concavity for $f(x) = \frac{x}{x^2 + 1}$.

Step 1: Find the first derivative (quotient rule)

$$f'(x) = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$$

Step 2: Find the second derivative (quotient rule again)

Let $u = 1 - x^2$ and $v = (x^2+1)^2$

$u' = -2x$

$v' = 2(x^2+1)(2x) = 4x(x^2+1)$

$$f''(x) = \frac{-2x(x^2+1)^2 - (1-x^2) \cdot 4x(x^2+1)}{(x^2+1)^4}$$

Factor out $2x(x^2+1)$ from numerator:

$$= \frac{2x(x^2+1)[-(x^2+1) - 2(1-x^2)]}{(x^2+1)^4} = \frac{2x[-x^2-1-2+2x^2]}{(x^2+1)^3} = \frac{2x(x^2 - 3)}{(x^2+1)^3}$$

Step 3: Find where $f''(x) = 0$

$2x(x^2 - 3) = 0$

$x = 0$ or $x = \pm\sqrt{3}$

Step 4: Test intervals (denominator is always positive)

Interval Sign of $2x$ Sign of $(x^2-3)$ $f''(x)$ Concavity
$(-\infty, -\sqrt{3})$ − + − Down
$(-\sqrt{3}, 0)$ − − + Up
$(0, \sqrt{3})$ + − − Down
$(\sqrt{3}, \infty)$ + + + Up

$$\boxed{\text{Concave up: } (-\sqrt{3}, 0) \cup (\sqrt{3}, \infty)}$$ $$\boxed{\text{Concave down: } (-\infty, -\sqrt{3}) \cup (0, \sqrt{3})}$$


Tips for Concavity Problems

  1. Second derivative = concavity — always find $f''(x)$ first
  2. Make a sign chart — organized tables help track sign changes
  3. Check both conditions for inflection points: $f''(x) = 0$ AND concavity changes
  4. Factor $f''(x)$ when possible — makes sign analysis easier
  5. Don't forget undefined points — if $f''(x)$ is undefined, check there too

Common Mistakes and Misunderstandings

❌ Mistake: Assuming $f''(x) = 0$ always gives an inflection point

Wrong: "$f''(0) = 0$ for $f(x) = x^4$, so $(0, 0)$ is an inflection point."

Why it's wrong: You must check that concavity actually changes. For $x^4$, it's concave up on both sides of $x = 0$.

Correct: Test intervals on both sides. If concavity doesn't change, it's NOT an inflection point.


❌ Mistake: Confusing increasing/decreasing with concave up/down

Wrong: "The function is going up, so it's concave up."

Why it's wrong: A function can be increasing while concave down (like the left half of $-x^2 + 10$ near the peak).

Correct: Increasing/decreasing is about $f'(x)$. Concavity is about $f''(x)$.


❌ Mistake: Forgetting to find the $y$-coordinate of inflection points

Wrong: "The inflection point is at $x = 2$."

Why it's wrong: An inflection point is a point on the curve, so you need both coordinates.

Correct: Plug $x = 2$ back into $f(x)$ to get the $y$-value. Report as $(2, f(2))$.

Formulas & Reference

Second Derivative Test for Concavity

$$f''(x) > 0 \implies \text{Concave Up} \qquad f''(x) < 0 \implies \text{Concave Down}$$

The sign of the second derivative tells you the concavity. Positive means the curve opens upward (like a bowl); negative means it opens downward (like a hill).

Variables:
$f''(x)$:
The second derivative of f
$\text{Concave Up}$:
Curve opens upward, tangent lines below curve
$\text{Concave Down}$:
Curve opens downward, tangent lines above curve
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