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Unfortunately, no. However $f'$ is severely constrained in this instance. It can be shown that $f'$ actually satisfies the conclusion of the intermediate value theorem. To give an example of this phenomenon, take $f(x) = x^2 \sin(\frac{1}{x})$ when $x\ne 0$, and $f(0) = 0$. Then $f$ is differentiable everywhere, but its derivative is not continuous at the ...
For continuity, the only point of concern is $x=0$ and we need $\lim_{x \uparrow 0} f(x) = 1 = \lim_{x \downarrow 0} f(x) = b$. Hence $f$ is continuous everywhere iff $b=1$. Since $f_-(x)= 2x^2+x+1$ and $f_+(x)= ax+b$ are both smooth everywhere, the only way that $f$ can fail to be differentiable at $x=0$ is if $f_-'(0) \neq f_+'(0)$. Hence we need $f_-'(... 4 On$[0,1]$we have that$\cos{x} \neq 0$and also it is continuous. The numerator is continuous as a product of the continuous function$e^{x^2}$and the composition of the continuous$\sqrt{x}$with$\sin{x}$(which is continuous) Ratio of continuous functions is continuous. 4$T= \{x \in \mathbb R \mid (f-g)(x)=0\} = (f-g)^{-1}(\{0\})$, i.e.$T$is the inverse image under$f-g$of the singleton$\{0\}$. As$\{0\}$is closed (a singleton set is closed) and$f-g$continuous,$T$is closed. 4 Yes, of course, because the integral of a positive continuous function is positive: $$\int_0^1g(x)dx-\int_0^1f(x)dx=\int_0^1(g(x)-f(x))dx>0.$$ 4 The set of points of discontinuity can be everywhere dense; in fact, it can be any countable subset of the plane. Let$D=\{(x_n,y_n):n\in\mathbb N\}$be a countable subset of$\mathbb R^2$. You can easily contruct a function$h:\mathbb R^2\to[0,1]$which is discontinuous at$(0,0)$and continuous everywhere else, and has the property that its restriction ... 3 Recall: a function$f$is continuous at$c$if $$\lim_{x \to c^+} f(x) = \lim_{x \to c^-} f(x)$$ If one graphs your proposed$f$(in red below, with$x=7$in blue), we can see this is not the case: one limit is is$+\infty$and the other is$-\infty$: The likely conclusion is there is a typo; I second Michael Hardy's suggestion from the comments in this ... 3 Let$(x_n)$be a convergent sequence in$T$with limit$x_0$. You have to show that$x_0 \in T$. To this end use$f(x_n) =g(x_n)$for all$n$,$x_0 \in [a,b],f(x_n) \to f(x_0)$and$g(x_n) \to g(x_0).$Can you proceed ? 3 No. Urysohn's lemma holds if and only if the space is normal. However not every locally compact Hausdorff space is normal. (Check here: https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal) Nevertheless, there is a version of Urysohn's lemma that holds in your case: if$\Omega$is a locally compact Hausdorff space and ... 3 No, there exist such functions that don't have a minimal period. Set$p_0=1$and choose$p_n \in \mathbb Q, n=1,2,\ldots$with $$p_1 > p_2 > \ldots > p_n > p_{n+1} > \ldots,\quad\lim_{n \to \infty}p_n=1=p_0.$$ Those$p_n, n=0,1,\ldots$will be the periods of a function that is very close to the final function$f$that will serve as example.... 3 Following your idea, without loss of generality, and towards a contradiction,$f(b)>f(a)$. Define$g(x)=\frac{f(x)-f(a)}{x-a}$if$x\neq a$and$g(a)=0.$Then$g$is continuous on$[a,b]$. Now, set$g(b)=2\epsilon$and$c=\inf\{x\in [a,b]:g(x)>\epsilon\}.$Then, continuity of$g$implies that$c < b$and$g(c)=\epsilon.$(Drawing a picture here ... 2 This is similar to the idea in Kavi Rama Murthy's answer. First, note that for$k = 0, 1, 2, \dots$, you can construct a function$h_k : [-1, 1] \to \mathbb{R}$which is$C^k$but is not$(k+1)$times differentiable at any point: just take repeated antiderivatives (integrals) of a continuous nowhere differentiable function, e.g. the Weierstrass function. By ... 2 You can redefine if the discontinuity is removable i.e. if the limit exists at 7. But here the limit does not exist. 2$h(x):=f(x)-g(x)$,$h$is continuous.$T=${$x| h(x)=0$}. Then$T= h^{-1}${$0$} is closed being the inverse image of the closed set {$0$}. 2 Suppose$y\in[a,b]\setminus T$. Then$a:=|f(y)-g(y)|\gt 0$. As$f$and$g$are continuous, we can find a$\delta\gt0$such that $$|f(x)-f(y)|\lt \frac a3,\:|g(x)-g(y)|\lt \frac a3$$whenever$|x-y|\lt\delta$. So if$|x-y|\lt\delta, by triangle inequaltiy, \begin{align*}a=|f(y)-g(y)|&\leq |f(y)-f(x)|+|f(x)-g(x)|+|g(x)-g(y)|\\&\lt\frac a3+|f(x)-g(x)|... 2 Yes. If A \subseteq \mathbb{R} is bounded, then A is contained in some compact set K \subseteq \mathbb{R}. The continuous image of a compact set is compact, and thus bounded. So we have f(A) \subseteq f(K) which is bounded, and thus f(A) is as well. 2 So you're considering X \times \{t\} as a subspace of X \times I in the induced topology, and you want to show that f: X \rightarrow X \times \{t\}, x \mapsto (x,t) is continuous. Not only is f continuous; it's a homeomorphism. Since the open sets in X \times I are unions of sets of the form U \times V with U open in X and V open in I, ... 2 Write your term as\frac{x^3}{y^2}\frac{x+y^2-x}{\sqrt{x}+\sqrt{y^2+x}}2 |\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )|=|\frac{x^3}{\sqrt{x}+\sqrt{y^2+x}}| \leq |x|^{\frac{5}{2}} So the limit of the function at zero is zero. 2 As I said in the comments: a^3 - b^3 = (a - b)(a^2 + ab + b^2) so we can write this as \begin{align} &3|(x^2 + 1) - (y^2 + 1)| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2| \\ ={} & 3|x - y| \cdot |x + y| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2|. \end{align} Then you just need a crude bound on 3|x + y| \cdot |(x^2 + 1)^2 + ... 2 In answer to your first question, yes, this is because of the unboundedness assumption. If no such x_n existed for some n \in \Bbb{N}, then this would imply that f(x) < n for all x \in [a, b], which is to say f is bounded above by n. In answer to your second question, this follows from the first highlight. You know that f(x_n) \ge n for all ... 2 You don't need to use the \varepsilon-\delta definition. First you should know that that the metric as a map d: M \times M \to \Bbb R is continuous, where M \times M has the product topology (or product metric, if you prefer), and \Bbb R the standard topology. This is a standard fact proven many times over on this site. Then (1,f): M \to (x, f(... 1 For an explicit construction of the type of function suggested in the comment above take n=1 and f(x)=\sum_k I_{(k-\frac 1 {k^{3}},k+\frac 1 {k^{3}}) }(x) (k^{4}|x-k|-k). This function is continuous and integrable but |f(k)|=k \to \infty. 1 Counterexample that's easy to write down:f(x) = x\left ( \dfrac{2+\sin x}{3} \right )^{x^5},$$although it requires some work to verify. 1 This is clearly continuous for ab\ne 0 so the question is how to define it on (0, b, c) and (a, 0, c). I would rewrite it as c\frac{ sin(ac)}{ac}\frac{1- cos(ab)}{(ab)^2}. 1 No, you must show that there is an open set O in the box topology so that f^{-1}[O] is not open (not the same as closed!). Try$$O= \prod_{n =1}^\infty (-\frac1n,\frac1n)$$and show that f^{-1}[O]=\{0\}, indeed not open. 1 A proof not using contradiction: Let A_n := \{x \in X \mid f(x) < n\}. As f is upper-semicontinuous, A_n is an open set. We have: X = \cup_{n \in \mathbb N} A_n. X is compact, therefore it has a finite subcover F \subset \mathbb N, i.e., X = \cup_{n \in F} A_n, which shows that f is bounded from above by M = \max_{n \in F} n. 1 Let \{q_1,q_2,....\} an enumeration of the rationals.Take the function$$f(x)=\sum_{q_n <x}\frac{1}{n^2}$$The functions is clearly bounded and increasing by density of rationals. Now let q_N \in \Bbb{Q} Then, for all x > q_N, note that$$f(x) = \sum_{ x > q_n} \frac 1{n^2} = \sum_{ q_N > q_n} \frac 1{n^2} + \sum_{ x > q_n \geq ... 1 We havef(x)$continuous if for all$a\in [0,1]$hold$\lim_{x\to a}f(x)=f(a). By limit proprerties we have \begin{align} \lim_{x\to a}f(x) =& \lim_{x\to a} \frac{e^{x^2}\cdot\sqrt{\sin(x)}}{\cos(x)} \\ =& \frac{\lim_{x\to a} e^{x^2}\cdot \lim_{x\to a} \sqrt{\sin(x)}}{\lim_{x\to a} \cos(x)} \\ =& \frac{ e^{\lim_{x\to a} x^2}\cdot \sqrt{\... 1 For any given\alpha>0$, choose an integer$nso large that $$n>\frac{1}{2\pi\alpha}.$$ Define \begin{align*} x_1&\equiv\frac{1}{2n\pi},\\x_2&\equiv\frac{1}{(2n+1)\pi}. \end{align*} Clearly,x_1$and$x_2$are both in$(0,\alpha)\$. Moreover, \begin{align*} f'(x_1)&=2x_1\sin\left(\frac{1}{x_1}\right)-\cos\left(\frac{1}{x_1}\right)+c\\ &...