52 Derivatives and the Intermediate Value Property

#todo/school : Problems from §5.2: 2, 3a, 4, 5, 7, 10 📅 2024-12-03 ✅ 2024-12-03

5.2.2

Question

Exactly one of the following requests is impossible. Decide which it is, and provide examples for the other three. In each case, let's assume the functions are defined on all of $R$ .
a. Functions $f, g$ not differentiable at zero but where $f g$ is differentiable at zero.
b. A function $f$ not differentiable at zero and a function $g$ differentiable at zero where $f g$ is differentiable at zero.
c. A function $f$ not differentiable at zero and a function $g$ differentiable at zero where $f + g$ is differentiable at zero.
d. A function $f$ differentiable at zero but not differentiable at any other point.

Proof

a. Possible. Use $f (x) = g (x) = \sqrt{| x |}$ . Here $(f g) (x) = x^{2}$ which is differentiable at zero but $f, g$ both are not.

b. Possible. Notice in comparison to (c) that that argument doesn't work if we can have $g^{'} (0) = 0$ . Thus use $f (x) = \frac{1}{x}$ and $g (x) = 0$ . Then clearly $f g = 0$ is differentiable at 0 (along with $g$ ), but $f$ is not differentiable at 0.

c. Impossible. If $f + g$ is differentiable at zero and $g$ is as well, then:

(f + g) - g = f

by the Algebraic Differentiability Theorem must be differentiable at zero.

d. Possible. Using Thomae's Function as a starting point we have:

t (x) = {\begin{cases} 1 & x = 0 \\ \frac{1}{n} & x = \frac{m}{n} \in Q - {0} is in lowest terms with n > 0 \\ 0 & x \notin Q \end{cases}

was continuous at every strictly irrational point. We want kind-of the opposite, because we need continuity at zero in order to get close to differentiability. If we define:

f (x) = x t (x)

(similar to our discussion in Lecture 29 - Chain Rule#Important Examples of Peculiarly Derived Functions). Notice that at any $c \neq 0$ either $c$ is rational (and thus $t$ is not continuous at $c$ , so the derivative cannot exist there), or it is irrational (in which case $c t (c)$ is continuous, but not differentiable due to different converging sequence, via Existence of Functional Limits). At $c = 0$ though we have:

lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = lim_{x \to 0} \frac{x t (x)}{x} = lim_{x \to 0} t (x) = 0

Thus $f^{'} (0) = 0$ and exists!

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5.2.3

Question

a. Use the definition of a derivative to produce the proper formula for the derivative of $h (x) = \frac{1}{x}$ .
b. Combine the result of (a) with the Chain Rule to supply a proof for (iv) of Algebraic Differentiability Theorem.
c. Supply a direct proof of Algebraic Differentiability Theorem (iv) by algebraically manipulating the different quotient for $(\frac{f}{g})$ in a style similar to the proof for (iii) for the ADT.

Proof

\begin{aligned} h^{'} (c) & = lim_{x \to c} \frac{h (x) - h (c)}{x - c} \\ = lim_{x \to c} \frac{\frac{1}{x} - \frac{1}{c}}{x - c} \\ = lim_{x \to c} \frac{c - x}{(x - c) (x c)} \\ = lim_{x \to c} \frac{- 1}{x c} \\ = \frac{- 1}{c} lim_{x \to c} \frac{1}{x} \\ = \frac{- 1}{c} \cdot \frac{1}{c} \\ = \frac{- 1}{c^{2}} \end{aligned}

Thus $h^{'} (x) = \frac{- 1}{x^{2}}$ .

b. We want to prove that ${(\frac{f}{g})}^{'}$ is differentiable at $c$ when $g^{'} (c) \neq 0$ and is:

{(\frac{f}{g})}^{'} (c) = \frac{f^{'} (c) g (c) - f (c) g^{'} (c)}{g (c)^{2}}

Now notice that:

(\frac{f}{g}) (x) = f (x) h (g (x))

So using the Chain Rule and the Algebraic Differentiability Theorem (iii) then:

\begin{aligned} {(\frac{f}{g})}^{'} (x) & = f (x) h^{'} (g (x)) g^{'} (x) + f^{'} (x) h (g (x)) \\ = f (x) \cdot \frac{- 1}{g^{2} (x)} \cdot g^{'} (x) + f^{'} (x) \cdot \frac{1}{g (x)} \\ = \frac{1}{g (x)^{2}} \cdot (f^{'} (x) g (x) - f (x) g^{'} (x)) \end{aligned}

which is exactly what we wanted!

c. We did this actually in the proof for Algebraic Differentiability Theorem.

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5.2.4

Question

Follow these steps to provide a slightly modified proof of the Chain Rule.
a. Show that a function $h : A \to R$ is differentiable at $a \in A$ iff $\exists l : A \to R$ which is continuous at $a$ and satisfies:

h (x) - h (a) = l (x) (x - a) \forall x \in A

b. Use this criterion for differentiability (in both directions) to prove the Chain Rule.

Proof

a.
( $\Rightarrow$ ): Suppose $h : A \to R$ is differentiable at $a$ . Construct $l$ as:

l (x) = \frac{h (x) - h (a)}{x - a}

Notice that since $h$ is differentiable at $a$ then:

lim_{x \to a} \frac{h (x) - h (a)}{x - a} = lim_{x \to a} l (x)

exists, and thus $l$ is continuous at $a$ .

( $\Leftarrow$ ): Suppose $\exists l : A \to R$ where $l$ is continuous at $a$ and defined such that:

h (x) - h (a) = l (x) (x - a)

for all $x \in A$ . We want to show that for our $a \in A$ that $h$ is differentiable, meaning we need to show the following exists:

lim_{x \to a} \frac{h (x) - h (a)}{x - a} = lim_{x \to a} \frac{l (x) (x - a)}{x - a} = lim_{x \to a} l (x)

which it does by $l$ being continuous.

b. Let $f : I \to R$ and $g : J \to R$ where $I, J$ are intervals and $f (I) \subseteq J$ . Suppose $f$ is differentiable at $c$ and $g$ is differentiable at $f (c)$ . We want to show that:

(g \circ f)^{'} (c) = g^{'} (f (c)) f^{'} (c)

To do this, we can try to use (a) to construct some $l$ that is continuous at $a$ instead. Notice first that if we multiply the top and bottom by $f (y) - f (c)$ , letting $y \in A$ be arbitrary:

\begin{aligned} (g \circ f)^{'} (c) & = lim_{y \to c} \frac{g (f (y)) - g (f (c))}{x - c} \\ = lim_{y \to c} \frac{g (f (y)) - g (f (c))}{f (x) - f (c)} \cdot \frac{f (x) - f (c)}{x - c} \\ = lim_{y \to c} \frac{g (f (y)) - g (f (c))}{f (x) - f (c)} \cdot f^{'} (c) \end{aligned}

via the Algebraic Limit Theorem for Functional Limits. But the limit exists via $g$ being differentiable at $c$ . Define $l (y) = \frac{g (y) - g (f (c))}{y - f (c)}$ . Notice that $l$ is continuous at $c$ since $l (c) = g^{'} (f (c))$ which exists. Further:

(g c i r c f)^{'} (c) = lim_{y \to c} l (y) \cdot f^{'} (c) = g^{'} (f (c)) \cdot f^{'} (c)

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5.2.5

Question

Let $f_{a} (x) = {\begin{cases} x^{a} & x > 0 \\ 0 & x \leq 0 \end{cases}$ .
a. For which values of $a$ is $f$ continuous at zero?
b. For which values of $a$ is $f$ differentiable at zero? In this case, is the derivative function continuous?
c. For which values of $a$ is $f$ twice-differentiable?

Proof

a. $f$ is continuous at zero iff:

\begin{aligned} lim_{x \to 0^{-}} f_{a} (x) & = lim_{x \to 0^{+}} f_{a} (x) \\ 0 & = lim_{x \to 0^{+}} x^{a} \end{aligned}

Thus if $a = 0$ then that doesn't work, and any $a < 0$ gives an asymptote that makes the limit not exist. All other values give the limit value we expect, so then it's continuous iff $a > 0$ .

b. $f$ is differentiable at zero iff:

\begin{aligned} lim_{x \to 0} \frac{f_{a} (x) - f_{a} (0)}{x - 0} & = lim_{x \to 0} \frac{f_{a} (x)}{x} \end{aligned}

exists. Considering both sides:

\begin{aligned} lim_{x \to 0^{-}} \frac{f_{a} (x)}{x} & = lim_{x \to 0^{+}} \frac{f_{a} (x)}{x} \\ lim_{x \to 0^{-}} \frac{0}{x} & = lim_{x \to 0^{+}} \frac{x^{a}}{x} \\ 0 & = lim_{x \to 0^{+}} x^{a - 1} \end{aligned}

Now similar to $a$ this only exists iff $a > 1$ . When this happens then:

f_{a}^{'} (x) = {\begin{cases} a x^{a - 1} & x > 0 \\ 0 & x \leq 0 \end{cases}

which is only continuous at all points $a > 1$ (repeat the (a) argument with a constant $a$ in the front).

c. Doing a similar process for (a) and (b), we get that $a > 2$ for it to be twice differentiable.

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5.2.7

Question

Let:

g_{a} (x) = {\begin{cases} x^{a} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}

Find a particular (potentially non-integer) value for $a$ so that:
a. $g_{a}$ is differentiable on $R$ but such that $g_{a}^{'}$ is unbounded on $[0, 1]$ .
b. $g_{a}$ is differentiable on $R$ with $g_{a}^{'}$ continuous but not differentiable at zero.
c. $g_{a}$ is differentiable on $R$ and $g_{a}^{'}$ is differentiable on $R$ but such that $g_{a}^{″}$ is not continuous at zero.

Proof

Before any parts, $g_{a}$ is differentiable on $R$ iff:

\begin{aligned} lim_{x \to 0} \frac{g_{a} (x) - g_{a} (0)}{x} & = lim_{x \to 0} \frac{x^{a} \sin (\frac{1}{x})}{x} \\ = lim_{x \to 0} x^{a - 1} \sin (\frac{1}{x}) \end{aligned}

exists. So for differentiation we need $a > 1$ here. For being unbounded, computing the derivative:

g_{a}^{'} (x) = {\begin{cases} a x^{a - 1} \sin (\frac{1}{x}) - x^{a - 2} \cos (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}

So for having an unbounded derivative, we need $a - 2 > 0 \Rightarrow a > 2$ .

a. Notice first that similar to 52 Derivatives and the Intermediate Value Property#5.2.5 that this exists (ie: it's differentiable) in the case that $a > 1$ , but also that that it is unbounded until $a < 2$ (see above). So use $a = 1.5$ .

b. If we want $g_{a}^{'}$ to be continuous but not differentiable at zero, we need differentiable, but unboundedness (similar to (a)) for $g_{a}^{'}$ now, so do the same processes we did in the scratch with $g_{a}^{'}$ instead. You'll get that $a > 2$ for the differentiability part, and $a < 3$ for the continuity part. Thus use $a = 2.5$ .

c. This is the exact same as (b), just use $g_{a}^{″}$ instead of $g_{a}^{'}$ . Thus you'll need $a > 3$ (differentiability) and $a < 4$ (unbounded, no continuity at 0) so use $a = 3.5$ .

Notice for this whole process that:

g_{a}^{'} (x) \sim a g_{a - 1} (x) - g_{a - 2} (x)

Hence we could've used this property instead, but that wouldn't be mathematically rigorous.

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5.2.10

Question

Recall that a function $f : (a, b) \to R$ is increasing on $(a, b)$ if $f (x) \leq f (y)$ for all $x < y \in (a, b)$ . A familiar mantra from calculus is that a differentiable function is increasing if its derivative is positive, but this statement requires some sharpening in order to be completely accurate. Show that the function:

g (x) = {\begin{cases} \frac{x}{2} + x^{2} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}

is differentiable on $R$ and satisfies $g^{'} (0) > 0$ . Now prove that $g$ is not increasing over any open interval containing 0.
In the next section we will see that $f$ is indeed increasing on $(a, b)$ iff $f^{'} (x) \geq 0$ for all $x \in (a, b)$ .

Proof

For comprehension, let's find the derivative numerically at 0, since at any point $x \neq 0$ it's just:

g^{'} (x) = \frac{1}{2} + 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})

For $x = 0$ :

\begin{aligned} g^{'} (0) & = lim_{x \to 0} \frac{g (x) - g (0)}{x - 0} \\ = lim_{x \to 0} \frac{\frac{x}{2} + x^{2} \sin (\frac{1}{x}) - 0}{x - 0} \\ = lim_{x \to 0} \frac{1}{2} + x \sin (\frac{1}{x}) \\ = \frac{1}{2} + 0 \\ = \frac{1}{2} \end{aligned}

Where the limit for $x \sin (\frac{1}{x})$ was calculated in Lecture 29 - Chain Rule#Important Examples of Peculiarly Derived Functions. Thus:

g^{'} (x) = {\begin{cases} \frac{1}{2} + 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x}) & x \neq 0 \\ \frac{1}{2} & x = 0 \end{cases}

Notice that this value equals $lim_{x \to 0} g^{'} (x)$ that we calculated above, so then $g$ is differentiable on all of $R$ . Further, we've shown that $g^{'} (0) = \frac{1}{2} > 0$ . But notice that $g^{'} (x) < 0$ at some point around zero, which the next step shows that $g$ is not increasing.

Now to show that $g$ is not increasing on any open interval containing $0$ , let $a, b \in R$ such that $a < 0 < b$ so that $(a, b)$ is our open interval containing 0. WLOG we'll show that we can always find a point $x \in (a, 0)$ such that $g (x) > g (0) = 0$ , thus contradicting increasing. Choose $x = \frac{1}{2 π n}$ such that $a < x = \frac{1}{2 π n}$ (which is guarunteed by the Archimedian Principle). Then:

\frac{x}{2} + x^{2} \sin (\frac{1}{x}) = π n + \frac{1}{4 π^{2} n^{2}} \sin (2 π n) = π n > 0

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