Lecture 24: Monotonicity, Convexity, and Inflection Points

1.1 Determining Monotonicity

1.1.1 Derivatives and Monotonicity

• Geometric illustration:
• 

Theorem: derivatives-and-monotonicity

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$.

1. If $f'(x)\ge0$, then $f$ is nondecreasing on $[a,b]$.
2. If $f'(x)\le0$, then $f$ is nonincreasing on $[a,b]$.
3. If $f'(x)>0$, then $f$ is strictly increasing; if $f'(x)<0$, then $f$ is strictly decreasing.

The condition $f'(x)\ge0$ together with only finitely many zeros of $f'$ is also sufficient for strict increase; the decreasing case is analogous. More generally, strict monotonicity follows if $f'$ is not identically zero on any subinterval.

1.1.2 Examples

Example 1: Determine the monotonic intervals of $f(x)=e^x-x-1$

• For $x<0$, $f'(x)<0$, so $f$ is strictly decreasing on $(-\infty,0)$.
• For $x>0$, $f'(x)>0$, so $f$ is strictly increasing on $(0,+\infty)$.

Example 2: Prove that $x-\frac{x^3}{6}<\sin x<x$ for $x>0$

• Let $F(x)=x-\sin x$. Since $F(0)=0$ and $F'(x)=1-\cos x\ge0$, we have $F(x)>0$ for $x>0$, hence $\sin x<x$.
• Let $G(x)=\sin x-x+\frac{x^3}{6}$. Then $G(0)=G'(0)=G''(0)=0$ and

Applying monotonicity successively gives $G''(x)>0$, $G'(x)>0$, and $G(x)>0$ for $x>0$. Therefore $x-\frac{x^3}{6}<\sin x$.

1.2 Convexity and Inflection Points

1.2.1 Definitions and the Second-Derivative Test

Definition: convexity

Let $f$ be continuous on an interval $I$. For any distinct $x_1,x_2\in I$:

1. If

then $f$ is strictly convex on $I$.

1. If the inequality is reversed, then $f$ is strictly concave on $I$.

Geometric meaning

• Convex: each chord lies above the graph, and the slope increases.
• 
• Concave: each chord lies below the graph, and the slope decreases.
• 

Theorem: second-derivative-test

Let $f$ be continuous on $[a,b]$ and twice differentiable on $(a,b)$.

• If $f''(x)>0$, then $f$ is convex on the interval.
• If $f''(x)<0$, then $f$ is concave on the interval.

The sign of the first derivative determines monotonicity; the sign of the second derivative determines convexity or concavity.

1.2.2 Inflection Points

Definition: inflection-point

An inflection point is a point on a continuous curve where the curve changes from convex to concave or from concave to convex.

Candidate abscissas satisfy one of the following:

1. $f''(x_0)=0$;
2. $f''(x_0)$ does not exist, but $f$ is continuous at $x_0$.

A candidate is not automatically an inflection point. The convexity must actually change across $x_0$.

1.2.3 Examples

Example 1: Determine the convexity of $y=x^3$

• On $(-\infty,0)$, $y''<0$, so the curve is concave.
• On $(0,+\infty)$, $y''>0$, so the curve is convex.
• Convexity changes across $x=0$, so $(0,0)$ is an inflection point.

• Original illustration:
• 

Example 2: Find the convexity intervals and inflection point of $h(x)=\sqrt[3]{x}$

For $x\ne0$,

• For $x<0$, $h''(x)>0$, so the curve is convex.
• For $x>0$, $h''(x)<0$, so the curve is concave.
• Although $h''(0)$ does not exist, $h$ is continuous at $0$ and the convexity changes there. Thus $(0,0)$ is an inflection point.

1.3 Common Question Types

Question Type 1: convexity-and-inflection-points

1. Compute $f''(x)$.
2. Find the points where $f''(x)=0$ or $f''(x)$ does not exist.
3. Use these candidates to divide the domain and determine the sign of $f''$ on each interval.
4. A candidate is an inflection point only if the convexity changes across it.

Question Type 2: number-of-roots

• Existence: if $f$ is continuous on $[a,b]$ and $f(a)f(b)<0$, the Intermediate Value Theorem guarantees at least one zero in $(a,b)$.
• Uniqueness: if $f$ is strictly monotone on an interval, it has at most one zero there.
• Rolle's theorem helps bound the number of roots: if $f$ has two distinct zeros, then $f'$ has at least one zero between them.

Question Type 3: proving-inequalities

• Move all terms to one side, define an auxiliary function, and use its derivative to establish monotonicity.
• The Lagrange Mean Value Theorem is useful for differences of function values.

For example, prove that for $x>0$,

Let $F(x)=\ln(1+x)-\frac{x}{1+x}$. Then

so $F(x)>0$. Next let $G(x)=x-\ln(1+x)$. Then

so $G(x)>0$. The desired inequality follows.