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Success is not final, failure is not fatal: It is the courage to continue that counts, Winston Churchill.
Extreme Value Theorem. If $f$ is a continuous function on a closed interval $[a, b]$, then $f$ has both an absolute maximum and an absolute minimum value on $[a, b]$.
Simply put, a continuous function on a closed interval cannot increase or decrease infinitely. Because it has boundaries and no breaks, it is guaranteed to hit a definitive highest point and lowest point (Figure 1.f).
Fermat’s Theorem. If a function $f(x)$ is defined on the interval $(a, b)$, has a local extremum (relative maximum or minimum) at $x = c$, and $f'(c)$ exists, then $x = c$ is a critical point of $f(x)$ and $f'(c) = 0$(Figure 1.a.).
Fermat’s theorem tells us where to look for extreme values. While the EVT guarantees these points exist, Fermat’s shows us that if an extremum occurs smoothly inside the interval, the tangent line at that peak or valley must be perfectly horizontal.
Intermediate Value Theorem (IVT). If a function $f$ is continuous on a closed interval $[a, b]$, then for every number $k$ between $f(a)$ and $f(b)$, there exists at least one point c within that interval $c \in [a, b]$ such that $f(c) = k$. For every number k between f(a) and f(b), there exist at least one point c within that interval $c \in [a, b]$ such that f(c) = k. In other words, it must take on every value between f(a) and f(b) at some point c within that interval.
Think of drawing the graph of a continuous function without lifting your pencil. If the function starts at height f(a) and ends at height f(b), then every height in between those two must be crossed somewhere along the way. The function cannot “jump over” any value between f(a) and f(b).
If $f(a)$ and $f(b)$ have opposite signs, $0$ must lie between them. Therefore, there must be some point $c$ where $f(c) = 0$.
Example: Does $f(x) = x^4 + x - 3$ have a root on the interval $(1, 2)$?
Narrowing Down the Root:
We can use a process of elimination to squeeze closer to the exact root:
Rolle’s Theorem. If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$ (Figure 1.c, 1.d).
If you start and end at the exact same height on a smooth curve, you must have leveled out (a stationary point somewhere between them where the slope is zero) at least once.
Proof:
Because $f$ is continuous on $[a, b]$, the Extreme Value Theorem states it must have an absolute maximum $M$ and minimum $m$.
Case 1: $M = m$
The function is perfectly flat (a constant line $f(x) = k$). Therefore, $f'(x) = 0$ for all $x$ in the interval, and any point $c$ works.
Case 2: $M \neq m$
Since the starting and ending heights are equal ($f(a) = f(b)$), the maximum and minimum cannot both occur at the endpoints.
At least one of them must occur at some point $c$ strictly inside the open interval $(a, b)$.
By Fermat’s Theorem, since an extremum occurs at $c$ and the function is differentiable there, $f'(c) = 0$.
The mean value theorem states that if $f$ is a continuous function on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c \in (a, b)$ such that the tangent at c is parallel to the secant line through the endpoints, that is, $f'(c) = \frac{f(b)-f(a)}{b-a}$. Figure 1.e. and 5 (at least a number $c \in (a, b)$), it may have two or more $c_1$ and $c_2$.
The MVT is the generalized version of Rolle’s Theorem. Geometrically, it means the slope of the secant line connecting the endpoints is exactly equal to the slope of the tangent line at some interior point $c$.
Physically, this means there is always a point where your instantaneous rate of change equals your average rate of change over the interval. (If you drive 60 miles in 1 hour, your speedometer had to read exactly 60 mph at least once).
Proof.
We prove this by tilting the graph so we can reuse Rolle’s Theorem! We define a new function $g(x)$ that represents the vertical distance between our curve $f(x)$ and the secant line.
Let $g(x) = f(x) - \left[ f(a) + \frac{f(b)-f(a)}{b-a}(x - a) \right]$
Because the secant line intersects $f(x)$ exactly at $x=a$ and $x=b$, the difference between them at those points is zero: $g(a) = 0$ and $g(b) = 0$.
Since $g(a) = g(b)$, Rolle’s Theorem applies to $g(x)$. There must be some point $c$ where $g'(c) = 0$.
Taking the derivative of $g(x)$ gives us: $g'(x) = f'(x) - \frac{f(b)-f(a)}{b-a}$
Set $g'(c) = 0$: $0 = f'(c) - \frac{f(b)-f(a)}{b-a} \implies f'(c) = \frac{f(b)-f(a)}{b-a}$
Therefore, $x = -1$ is the exact point where the tangent line is completely horizontal, making it perfectly parallel to the secant line connecting the endpoints.
Let $f(x) = \sin(x) + 2$ over the interval $[0, 2\pi]$.
To find them, we take the derivative and set it to zero: $f'(x) = \cos(x)$.
$\cos(c) = 0 \implies c = \frac{\pi}{2}, \quad c = \frac{3\pi}{2}$
Let $f(x) = \frac{1}{x^2}$ over $[-1, 1]$.
Endpoints: $f(-1) = 1$ and $f(1) = 1$.
You might think Rolle’s theorem guarantees a point where the derivative is zero since the endpoint values match. However, $f(x)$ is undefined at $x = 0$. Because the function is not continuous on the closed interval $[-1, 1]$, Rolle’s theorem does not apply.
The Mean Value Theorem states that for a continuous and differentiable curve, there is at least one point where the instantaneous rate of change (the tangent) equals the average rate of change (the secant line connecting the endpoints).
Let $f(x) = x^3 + 4x$ over $[-1, 1]$ (Figure 1).
$f'(x) = 3x^2 + 4$. Therefore, $3c^2 + 4 = 5 \implies 3c^2 = 1 \implies c^2 = \frac{1}{3} \implies c = \pm\frac{\sqrt{3}}{3} \approx \pm 0.58$. Both points lie within the valid interval $(-1, 1)$
Let $f(x) = \vert{}x\vert{} - 1$ over $[-1, 1]$.
Average Rate of Change: $f(-1) = 0$ and $f(1) = 0$. The average slope is $0$.
However, there is no point $c \in (-1, 1)$ where $f'(c) = 0$. Why? The absolute value function has a sharp corner (a “cusp”) at $x = 0$, meaning it is not differentiable at $x = 0$. If a function is not differentiable even at a single point within the interval, the theorem cannot guarantee a zero derivative (that is, the result may not hold).
Let $f(x) = -x^2 + 3x + 10$ over $[2, 6]$ (Figure 4).
To find where the instantaneous velocity matches this average, we set $f'(c) = -5$. We calculate the derivative, $f'(x) = -2x + 3$
$-2c + 3 = -5 \implies -2c = -8 \implies c = 4$
At exactly $x = 4$, the instantaneous rate of change equals the average rate of change.
A hot air balloon is launched into the air with a human pilot with a bad attitude. The twice-differentiable function f models the balloon’s height, measured in meters, at time t. The below table gives values of the function at selected times t.
| t (minutes) | 0 | 5 | 15 | 20 | 30 |
|---|---|---|---|---|---|
| f(t) (meters) | 0 | 40 | 70 | 65 | 80 |
Is there a time between $t = 20$ and $t = 30$ when the balloon’s velocity is exactly $1.5$ meters per minute?
$\text{Average Velocity on } (20, 30) = \frac{f(30) - f(20)}{30 - 20} = \frac{80 - 65}{10} = 1.5 \text{ meters/minute}
Because $f$ is twice-differentiable (which implies it is continuous and differentiable), the Mean Value Theorem guarantees that there must be at least one value $c \in (20, 30)$ where $f'(c) = 1.5$. We don’t need the exact formula of the function to prove this!
Let $f$ be continuous on the interval $[a, b]$ and differentiable on $(a, b)$. The sign of the derivative tells us the direction of the function:
Proof via MVT:
For any $x_1, x_2 \in (a, b)$ where $x_1 < x_2$, the difference $x_2 - x_1 > 0$.
The Mean Value Theorem guarantees we can find a $c \in (x_1, x_2)$ such that: $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$
If $f'(x) > 0$, then $f'(c) > 0$. Since $x_2 - x_1$ is also positive, $f(x_2) - f(x_1) > 0$, which means $f(x_2) > f(x_1)$. Therefore, the function is increasing.
Show that $f(x) = x^3 - 3x^2 + 3x + 5$ is strictly increasing on $\mathbb{R}$.
Because it is a polynomial, $f$ is continuous and differentiable everywhere.
$f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$
Since a squared term is always non-negative, $3(x - 1)^2 \geq 0$ for all $x \in \mathbb{R}$.
Thus, $f$ is increasing on $\mathbb{R}$.
Find the increasing/decreasing intervals for $f(x) = 2x^3 - 15x^2 + 36x + 5$.
Take the derivative and factor to find critical points: $f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$
Testing intervals around the critical points $x=2$ and $x=3$:
$\star \quad x < 2$: $f'(x) > 0 \implies f$ is increasing on $(-\infty, 2)$.
$\star \quad 2 < x < 3$: $f'(x) < 0 \implies f$ is decreasing on $(2, 3)$.
$\star \quad x > 3$: $f'(x) > 0 \implies f$ is increasing on $(3, \infty)$.
We can use derivatives to prove fundamental inequalities regarding the exponential function.
Let $f(x) = e^x - x - 1$.
$f(0) = e^0 - 0 - 1 = 0$.
$f'(x) = e^x - 1$.
Because $e^x$ is monotonically increasing and $e^0 = 1$, $f'(x) > 0$ for all $x > 0$. Therefore, $f$ is increasing for $x > 0$.
Since $f(0) = 0$ and $f$ is increasing, $f(x) > 0$ for all $x > 0$.
Therefore, $e^x - x - 1 > 0 \implies \boxed{e^x > x + 1}$
Let $g(x) = e^x - x - 1 - \frac{x^2}{2}$.
$g(0) = 0$.
$g'(x) = e^x - 1 - x$.
From our previous proof, we know $e^x - x - 1 > 0$, meaning $g'(x) > 0$ for all $x > 0$.
Since $g'(x)$ is positive, $g$ is increasing.
Therefore, $g(x) > 0 \implies \boxed{e^x > x + 1 + \frac{x^2}{2}}$
These inequalities are not random. The expressions $1 + x$ and $1 + x + \frac{x^2}{2}$ are the first and second-degree Taylor polynomials (Maclaurin series) for $e^x$. If you continue this process, you get $e^x > 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
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