The Mean Value Theorem and curve sketching

This week you learn “what the derivative says about the shape of a function”. The theoretical part is the Mean Value Theorem: it is what guarantees that you can extract information about a function from its derivative. The practical part is curve sketching: you read increasing/decreasing from the sign of $f'$ and concavity from the sign of $f''$, and draw the graph. There is almost always a “sketch the graph” question on the final, worth 15-20 points. The main ideas:

1. MVT: on an interval, at at least one point the tangent slope equals the secant slope.
2. The sign of $f'$ gives increasing/decreasing, and a sign change gives a local extremum.
3. The sign of $f''$ gives concavity, and a sign change gives an inflection point.
4. $f'(c)=0$ or $f''(c)=0$ alone proves nothing. Look at the sign.

Rolle’s Theorem

Let $f$ satisfy three conditions:

1. continuous on $[a,b]$,
2. differentiable on $(a,b)$,
3. $f(a)=f(b)$.

Then there is at least one $c\in(a,b)$ with $f'(c)=0$.

Intuition: a smooth path that starts and ends at the same height must make a peak or a valley somewhere in between. There the tangent is horizontal.

The Mean Value Theorem (MVT)

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$:

$\text{there is a } c\in(a,b) \text{ with } f'(c)=\frac{f(b)-f(a)}{b-a}.$

The right side is the slope of the secant connecting the endpoints. The theorem says that somewhere the tangent is parallel to this secant. Physical interpretation: if you traveled 180 km in 2 hours, at some moment your speedometer showed exactly 90 km/h.

Example 1. $f(x)=x^2+2x$, $[1,4]$.

secant slope = [f(4) - f(1)] / (4 - 1) = (24 - 3) / 3 = 7
f'(c) = 2c + 2 = 7   ->   c = 2.5     (1 < 2.5 < 4, valid)

PITFALL · applying it without checking the conditions. For $f(x)=|x|$ on $[-1,1]$, $f(-1)=f(1)$ but there is no point with $f'(c)=0$. Why? It is not differentiable at $0$, so the second condition fails. If the exam says “verify the hypotheses”, write out continuity and differentiability explicitly.

Consequences of the MVT

Consequence 1. If $f'(x)=0$ everywhere on $(a,b)$, then $f$ is constant there. (MVT between any two points: $f(x_2)-f(x_1)=0\cdot(x_2-x_1)=0$.)

Consequence 2. If $f'=g'$, then $f=g+C$. This is why there is a "$+C$" in integration.

Example 2 (proving an identity). For $x>0$, $\arctan x+\arctan\dfrac1x=\dfrac{\pi}{2}$.

Let $h(x)=\arctan x+\arctan\frac1x$. Derivative:

$h'(x)=\frac{1}{1+x^2}+\frac{1}{1+\frac{1}{x^2}}\cdot\left(-\frac{1}{x^2}\right)=\frac{1}{1+x^2}-\frac{1}{x^2+1}=0$

$h$ is constant. At $x=1$, $h(1)=\tfrac\pi4+\tfrac\pi4=\tfrac\pi2$. So $h(x)=\tfrac\pi2$ for every $x>0$.

Example 3 (inequality). For $x>0$, $\ln(1+x)<x$.

MVT for $f(t)=\ln(1+t)$ on $[0,x]$: $\ln(1+x)-\ln1=\dfrac{1}{1+c}\cdot x$, $0<c<x$. Since $\dfrac{1}{1+c}<1$, $\ln(1+x)<x$.

The increasing / decreasing test and the first derivative test

• If $f'>0$ on an interval, $f$ is increasing there.
• If $f'<0$, it is decreasing.

First derivative test: let $c$ be a critical point.

To build a sign table: place the critical points and undefined points on the number line, choose a test point in each interval, and look at the sign of $f'$.

Concavity and the second derivative

• $f''>0$: the graph is concave up, shaped like $\cup$. The tangents are below the graph.
• $f''<0$: concave down, shaped like $\cap$.
• Inflection point: a point where $f$ is defined and the concavity changes.

Second derivative test: let $f'(c)=0$. If $f''(c)>0$ it is a local min, if $f''(c)<0$ a local max, and if $f''(c)=0$ the test gives no result (go back to the first derivative test).

PITFALL · $f''(c)=0$ does not mean an inflection point. For $f(x)=x^4$, $f''(0)=0$ but $f''=12x^2\ge0$ everywhere, so the sign does not change. There is no inflection point at $0$; there is a minimum. For an inflection point, a sign change is required.

The curve sketching template

A. Domain
B. Intercepts: x = 0 -> y-intercept;  f(x) = 0 -> x-intercepts
C. Symmetry: even / odd / periodic
D. Asymptotes: vertical, horizontal, oblique
E. f': critical points, intervals of increase/decrease, local max/min
F. f'': intervals of concavity, inflection points
G. Mark the important points, draw following the asymptotes

Example 4 (a full sketch). $f(x)=\dfrac{x^2}{x-1}$.

A. $x\ne1$.

B. $f(0)=0$; $f(x)=0 \iff x=0$. The graph passes through the origin.

C. $f(-x)\ne\pm f(x)$, no symmetry.

D. $x\to1^+$: $\tfrac{1}{0^+}\to+\infty$; $x\to1^-$: $-\infty$. Vertical asymptote $x=1$. The numerator’s degree is one more; polynomial division: $x^2=(x-1)(x+1)+1$, so $f(x)=x+1+\dfrac{1}{x-1}$. Oblique asymptote $y=x+1$.

E. $f'(x)=\dfrac{2x(x-1)-x^2}{(x-1)^2}=\dfrac{x(x-2)}{(x-1)^2}$. The critical points are $0$ and $2$.

At $x=0$ there is a local max $f(0)=0$. At $x=2$ there is a local min $f(2)=4$.

F. From the form $f(x)=x+1+(x-1)^{-1}$: $f''(x)=2(x-1)^{-3}=\dfrac{2}{(x-1)^3}$. $x<1$: concave down; $x>1$: concave up. The sign changes at $x=1$ but that point is outside the domain: there is no inflection point.

G. On the left branch the graph comes in from $(-\infty)$ below the oblique asymptote, peaks at $(0,0)$, and drops to $-\infty$ as it approaches $x=1$. On the right branch it comes down from $+\infty$, bottoms out at $(2,4)$, then rises approaching $y=x+1$ from above.

TIP · take the derivative in a simple form. Finding $f''$ from $\dfrac{x(x-2)}{(x-1)^2}$ with the quotient rule is long and error-prone. The form $x+1+\dfrac{1}{x-1}$ from polynomial division finishes it in two lines.

Common mistakes