Calculus Pop Quiz

The correct statements are 1, 3, 4.

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Subparagraph before the solution:

Differentiable function is not the same as derivative function, at all. If $f:\mathbb{R}^n \to \mathbb{R}$ is a differentiable function then it has directionable derivates ,nevertheless a function is able to have a derivate on every direction and it has not to be a differentiable function. Over the compex numbers also are different concepts,...

However, in this case $f:\mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function if and only if it is a derivative function.

Proof

$\Rightarrow$ ) Let's suppose that $f:\mathbb{R} \to \mathbb{R}$ is a differentiable function, then $\forall c \in \mathbb{R}$ $f(x) - f(c) = df_{c} (x - c) + |x - c|\rho(x - c)$ $\forall x$ in a reduced neighborhood of c(neighborhood without c), where $df_{c}$ is a $\mathbb{R}$ -linear transformation and $\rho(h)$ is a function defined in a reduced neighborhood of c fulfilling $\rho(h) \to 0$ as $h \to 0$ which implies that $f(x) - f(c) = (x - c)df_{c} (1) + |x - c|\rho(x - c) \Rightarrow \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = df_{c}(1)...$

$\Leftarrow$ ) Let's suppose that $f:\mathbb{R} \to \mathbb{R}$ is a derivative function, then $\forall c \in \mathbb{R}$ $\lim_{x \to c} \frac{ f(x) - f(c)}{x - c} = f '(c) \Rightarrow$ $\lim_{x \to c} \frac{ f(x) - f(c)}{x - c} - f '(c) =0 \Rightarrow f(x) - f(c) - f '(c)(x - c) = o(|x - c|) \Rightarrow$ $f(x) = f(c) + f '(c)(x - c) + o (|x - c|)$ in a reduced neighborhood of c.... Read note below $\square$ .

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Solution

Consider a differentiable function $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 0$ and $f '(0) = 1$ .

$\boxed{1}$ .- $f(x) = f(0) + f '(0)x + o(|x|)$ in a neighborhood of 0, which implies that $\exists q > 0 \space / f(x) \approx 0 +1 \cdot x \approx x \space \forall x \in (0,q) \Rightarrow f(x) > 0 \space \forall x \in (0,q)$ .

$\boxed{2}$ .- The stament 2 and 5 are false . Let $f(x) = x - 2x^2\sin(\frac{1}{x}) \text{, if x is distinct to 0 and, } f(0) = 0$ . This function fulfills $f(0) = 0, f '(0) = 1$ is differentiable over $\mathbb{R}$ with not continuous derivate(it is not continuous on 0) and not fulfilling the statement 2 because in all neighborhood of 0 there is a point x where $f '(x) < 0 \Rightarrow$ the function is not increasing in any neighborhood of 0.

This example is also interesting because if $f:\mathbb{R}^n \to \mathbb{R}$ has partial derivates functions being continuous then f is a differentiable function. And this example shows that the reciprocal is not true.

$\boxed{3}$ .- The absolute value function is a continuous function and the composition of two continuous functions is a continuous function. Every differentiable function $f:\mathbb{R} \to \mathbb{R}$ is a continuous function: $\forall a \in \mathbb{R}, \lim_{x \to a} f(x) = \lim_{x \to a} f(a) + f ' (a)(x -a) + o (|x -a|) = f(a) \Rightarrow$ $\space \forall \epsilon > 0, \exists \delta > 0 \space / \space |f(x) - f(a)| < \epsilon \text{ if } 0 \leq |x - a| < \delta$

$\boxed{4}$ .- All continuous function is Riemann integrable, and due to Fundamental Calculus Theorem $h(x) = \int_{0}^{x} f(t) dt$ fulfills $h ' (x) =f(x)$ . Take $g(x)=\int_{0}^{x} h(t) dt \Rightarrow g '(x) = h(x) \Rightarrow g ''(x) = h '(x) = f(x)$

Note .- $o(|x|) = |x|o(1) = |x|\rho(x)$ where $\rho(x)$ is a function with $\rho(x) \to 0$ as $x \to 0$ ( $o(|x|)$ means $\lim_{x \to 0} \frac{o(|x|)}{x} = 0$ )

Request for @Calvin Lin : Our Compañero @Guillermo Templado posted a wonderful solution as a comment, but unfortunately that solution was deleted, along with the comments. Is there any way to bring Guillermo's comment back as a solution, with his consent? It was a long post, and it would be a shame for all that careful work to be lost.

considering the function to be y=(sinx) then we will not get the first statement true.