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Write the desired conclusion as $$ f(x) + (x-a) f'(x) = 0 $$ for some $x \in (a, b)$, and note that the left-hand side is the derivative of $(x-a)f(x)$. This suggests to apply Rolle's theorem to $g(x …

(iv) is not the same question as (iii). You have to assume that the function $f:[0, \infty) \to (0, \infty)$ satisfies $$ \tag {*} \exp\left(\frac 1y\int_0^y\ln f(x) \, dx\right) = \sqrt{f(y)} $$ for …

$g(x) := f(x) + x^n$ is continuous on $\Bbb R$, and for $x \ne 0$, $$ g(x) = f(x) + x^n = \bigl( \frac {f(x)}{x^n} + 1 \big) x^n \, . $$ Since $f(x)/x^n \to 0$ for $x \to \pm \infty$, and $n$ is even …

(Unless I made some error, the statement actually holds with $(-3, 3)$ replaced by $(-a, a)$ for any $a > 1/\sqrt 2$.) Without loss of generality we can assume that $$ g(0) \ge 0 \text{ and } g'(0) …

If $(a_n)$ is bounded above by $M$ then for all $m > n \ge 1$ $$ M \ge a_m = a_n + \sum_{k=n}^{m-1}(a_{k+1}-a_k) \ge a_n + (m-n) (a_{n+1}-a_n) \\ \implies a_{n+1}-a_n \le \frac{M - a_n}{m-n} \, . $$ …

The given solution is written a bit sloppy. Actually the mean value theorem is applied to the closed interval $[0, x]$ (on which $f$ is continuous and differentiable). The MVT states that $$ f(x)-f(0) …

If $f$ is increasing and differentiable on $[a,b]$ then $$ \frac{f(y)-f(x)}{y-x} \ge 0 $$ for all $x, y \in [a,b]$ with $x \ne y$, and therefore $$ f'(x) = \lim_{y \to x} \frac{f(y)-f(x)}{y-x} \ge 0 …

Hint: Induction is not needed here. Write $$ 1 = f(1) - f(0) = \sum_{k=1}^n f\left(\frac{k}{n}\right)- f\left(\frac{k-1}{n}\right) $$ and then apply the mean-value theorem to $f$ on each interval $(\f …

The statement is wrong, a counterexample is $f(x) = x^2-1/3$ on $[a, b]= [-1, 1]$. The condition $ \int_{-1}^1 f(x) \, dx = 0$ is satisfied, but for all $\xi, \eta \in (-1, 1)$ is $$ f'(\xi)(\xi-a)+f' …

A differentiable function is convex if and only if $f'$ is increasing, therefore we can formulate the problem as follows: Let $f:(0, \infty) \to \Bbb R$ be differentiable. Then $f'$ is increasing if …

The statement is true if $f'(t)$ is nowhere zero: If $f(0) = f(1)$ then $f(0) - f(1)$ is a scalar multiple of $f'(t_0)$ for any $t_0$. Otherwise choose a non-zero vector $v$ which is orthogonal to …

The notation $\{a_n\} \downarrow 0$ usually means that the sequence is non-increasing (and tends to zero). The Cauchy condensation test does not work without the monotony condition: If you define $a_n …

In order to show that $f$ is differentiable with $|f'(x)| = 1$ you need to show that $\frac{f(x)-f(y)}{x-y}$ can only take one of the values $+1$ or $-1$, but not both. This can be done as follows: Fo …

Denote the partial sums with $s_n = \sum^{n}_{k=1}k^2a_k$ and apply the Stolz–Cesàro theorem: $$ \lim_{n \to \infty} \frac{s_n}{n^3} = \lim_{n \to \infty} \frac{s_n - s_{n-1}}{n^3 - (n-1)^3} $$ hold …

Using the well known inequality $\log(x) \ge 1-1/x$ we get that $$ x \log(x) - x + 1 \ge x \left( 1-\frac 1x \right) - x + 1 = 0 $$ for all $x > 0$. This proves that $h'(x) \ge 0$, i.e. that $h$ is …