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4

If $a=-1$,$b=1$ and $f(x)=x^2$ then there isn't a c in the open interval as the maximum is at the end points.


3

The function $f$ considered as a map $$f:\quad{\mathbb R}\to{\mathbb R},\qquad x\mapsto y=f(x)$$ multiplies all distances by $2$. Therefore, if your $y$-tolerance is $\epsilon$, you should make sure that the allowed $x$-deviation is $\leq{\epsilon\over2}$.


2

$$\inf_{x,y} f(x,y)\leq \inf_{x}f(x,y)$$ and thus $$\inf_{x,y}f(x,y)=\liminf_{y\to\infty }\inf_{x,y}f(x,y)\leq \liminf_{y\to\infty }\inf_x f(x,y)$$ what conclude the proof.


2

Since $a+b>0$, it is particularly not zero. The only way for a product of two reals to give zero is if at least one of them is zero. Since $a+b\neq 0$, but $0 = (a-b)(a+b)$, it must be the case that $a-b=0$, i.e. $a=b$.


2

For a given positive integer $n$, let $p_i(n) = (a_1, a_2, \ldots, a_k)$ be a given partition of $n$ that satisfies the criteria. For each such partition $p_i(n)$, how many ways are there to generate a unique partition $p_i(n+1)$? Is there a bijection?


2

Functional dependence of $k$ given functions $F_i:\>\Omega\to{\mathbb R}$ $\>(1\leq i\leq k)$ with common domain $\Omega\subset{\mathbb R}^n$ means, intuitively, that there is a nontrivial function $$g:\quad{\mathbb R}^k\to{\mathbb R},\qquad y\mapsto g(y)$$ such that $$g\bigl(F_1(x),F_2(x),\ldots, F_k(x)\bigr)\equiv 0\qquad \forall x\in\Omega\ ...


1

Functional independence between two functions means that the level set of each of the functions intersects transversely the level set of the other function. In the plane, this means that a function $f$ functionally independent of another function $g$ cannot be written as $f=F(g)$ where $F$ is another function, because in this case, the level sets would be ...


1

HINT: The conditions allow $s_n$ to be negative.


1

HINT: As an alternative to an inductive approach: for each possible choice of $k$ (the number of terms), there is exactly one choice of $a_1$ that allows the inequalities to be satisfied and the sum of the $a_i$’s to be $n$.


1

Since $a+b \neq 0$ we can multiply $(a-b)(a+b) =0$ across by ${1 \over a+b}$ to get $a-b=0$, that is, $a=b$.


1

Since $f'$ has limit $l$ at $x_0$, then given $\epsilon>0$, $\exists \delta>0$, s.t. $|x-x_0|<\delta\implies |f'(x)-l|<\epsilon$ Consider such $x$,, by MVT($f$ differentiable on open interval, continuous on closed interval), we have $$\frac{f(x)-f(x_0)}{x-x_0}=f'(\xi)$$ where $\xi$ lies between $x,x_0$. Then by triangle inequality ...


1

(Originally this was a set of hints $\ldots$) The set $M$ is bounded by four parabolic arcs. One could compute its area by cutting it up into simpler shapes, each of them bounded by segments parallel to the axes and at most one arc. Instead another procedure is suggested: One can write the definition of $M$ in the form ...


1

Note $$|f(x) - f(1)| = |(3 - 2x) - (3 - 2)| = |2 - 2x| = 2|x - 1|.$$ So $|f(x) - f(1)| < \varepsilon$ if $|x - 1| < \frac{\varepsilon}{2}$. If you choose $\delta > \frac{\varepsilon}{2}$, then set $x = 1 + \frac{\varepsilon}{2}$. We have $|x - 1| = \frac{\varepsilon}{2} < \delta$, but $|f(x) - f(1)| = 2|x - 1| = \varepsilon$. Hence, the largest ...


1

I am following your idea. $$A = \{a \le x < x_0 | f(x) = 0 \}$$ $$B = \{x_0 < x \le b | f(x) = 0 \}$$ Let $c = \sup A$ and let $d = \inf B$ ($\sup B$ doesn't work here, since it's always $b$). Note $c<x_0<d$ Step 1: prove $f(x) > 0$ for $x \in (c, d)$ we will use contradiction. Assume $f(x) < 0$, for some $c<x<d$, by IVT ...



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