Mean Value Theorem

Sign of the Derivative Tells us about Monotonicity

Let

f

be differentiable on

[a, b]

then if for all

x \in (a, b)

we have:

$f^{'} (x) = 0$ then $f$ is constant
$f^{'} (x) > 0$ then $f$ is strictly increasing
$f^{'} (x) < 0$ then $f$ is strictly decreasing
$f^{'} (x) \geq 0$ then $f$ is increasing
$f^{'} (x) \leq 0$ then $f$ is decreasing

Matching Derivatives Implies Differing by a Constant

Suppose that

f, g : ℝ \to ℝ

and that

f^{'} (x) = g^{'} (x)

for all

x \in (a, b)

then show that

g (x) = f (x) + c

for some

c \in ℝ

Observe that we have

f^{'} (x) - g^{'} (x) = 0

and therefore

f - g

is constant so that there is some

c \in ℝ

such that

f (x) - g (x) = c

so we have

f (x) = g (x) + c

as needed.

Non-Zero Derivative Implies Locally One-to-One

Suppose that

f : (a, b) \to ℝ

has continuous derivative on

(a, b)

. If

f^{'} (x_{0}) \neq 0

prove that there is an interval

(a, c)

containing

x_{0}

such that

f

is one to one on

(a, c)

Without loss of generality assume that

f (x_{0}) > 0

and recall that

ℝ^{+}

is an open set, therefore since

f^{'}

is continuous then we know that

V : = {(f^{'})}^{- 1} (ℝ^{+})

is open and since

f^{'} (x_{0}) > 0

then

x_{0} \in V

therefore there exists some

x_{0} \in (c, d) \subseteq V = {(f^{'})}^{- 1} (ℝ^{+})

therefore

f^{'} ((c, d)) \subseteq ℝ^{+}

so that

f

is strictly increasing on

(c, d)

so that it is injective, as needed.

Rolles and Zeros

Suppose

f : [a, b] \to ℝ

and

f

is differentiable on

(a, b)

, suppose that

f

has

n \in ℕ_{2}

zeros in the interval

[a, b]

then

f^{'}

has at least

n - 1

zeros on

(a, b)

Nth Derivative has a Zero Condition

Suppose that

f \in C^{n}

and that

f

has

n + 1

zeros, then f ^ \left( n \right) ParseError: Got function '\left' with no arguments as superscript at position 5: f ^ \̲l̲e̲f̲t̲( n \right) has a zero

Induction using the previous corollary.

Forced Limit of the Derivative

Suppose that

f

is continuous on the interval

[a, b]

and is differentiable at all points of

(a, b)

except possibly at the single point

x_{0} \in (a, b)

. Show that if

\lim_{x \to x_{0}} f^{'} (x)

exists (and denote its value by

L

) , show that

f^{'} (x_{0})

exists and

f^{'} (x_{0}) = \lim_{x \to x_{0}} f^{'} (x)

In order to show that $f^{'} (x_{0})$ exists we can show that the right and left limits exist and are equal.

Consider $\lim_{h \to 0^{+}} \frac{f (x_{0} + h) - f (x_{0})}{h}$ , for any $h \in ℝ^{+}$ such that $x_{0} + h < b$ we consider the interval $[x_{0}, x_{0} + h]$ since $f$ is continuous here, then there exists some $c_{h} \in (x_{0}, x_{0} + h)$ such that $f^{'} (c_{h}) = \frac{f (x_{0} + h) - f (x)}{h}$ Note that as $h \to 0$ then $c_{0} \to x_{0}$ therfore we have $L = \lim_{c_{h} \to x_{0}} f' (c) = \lim_{h \to 0^{+}} \frac{f (x_{0} + h) - f (x)}{h}$ we can do the same thing from the left to show that both sides of the limit are equal and both equal $L$ as needed.