IVT, MVT & Extreme Value Theorem

$f(x) = x^5 - 3x + 1$.

$f$ is a polynomial, so it is continuous on $[0, 1]$.

$f(0) = 1 > 0$ and $f(1) = 1 - 3 + 1 = -1 < 0$.

$0$ is between $f(0)$ and $f(1)$, so by the IVT there is $c \in (0, 1)$ with $f(c) = 0$.

$f(x) = 2\cos x - x$.

$\cos x$ and $x$ are continuous everywhere, so $f$ is continuous on $\left[0, \tfrac{\pi}{2}\right]$.

$f(0) = 2 > 0$ and $f\!\left(\tfrac{\pi}{2}\right) = 0 - \tfrac{\pi}{2} < 0$.

By the IVT there is $c \in \left(0, \tfrac{\pi}{2}\right)$ with $f(c) = 0$, that is $2\cos c = c$.

$f(x) = \log x + x - 3$, defined for $x > 0$.

$\log x$ and $x - 3$ are continuous for $x > 0$, so $f$ is continuous on $[1, e]$.

$f(1) = 0 + 1 - 3 = -2 < 0$ and $f(e) = 1 + e - 3 = e - 2 > 0$.

By the IVT there is $c \in (1, e)$ with $\log c = 3 - c$.

$f(x) = x^3 + 2x - 5$ is continuous; $f(1) = -2 < 0$, $f(2) = 7 > 0$, so by the IVT a root lies in $(1, 2)$.

$f'(x) = 3x^2 + 2$.

$3x^2 \ge 0$, so $f'(x) \ge 2 > 0$ for every real $x$.

$f$ is strictly increasing on $\mathbb{R}$, so it crosses $0$ at most once. With existence: exactly one.

$f(x) = e^{x} + x - 1$. A root is visible: $f(0) = 1 + 0 - 1 = 0$.

$f'(x) = e^{x} + 1$.

$e^{x} > 0$, so $f'(x) > 1 > 0$ for every $x$.

Suppose $f(a) = f(b) = 0$ with $a < b$. $f$ is differentiable, so Rolle gives $c \in (a, b)$ with $f'(c) = 0$ — impossible. So $x = 0$ is the only root.

$f(x) = x^4 + 4x - 1$ is continuous; $f(0) = -1 < 0$, $f(1) = 4 > 0$, so by the IVT a root lies in $(0, 1)$.

$f'(x) = 4x^3 + 4$.

For $0 \le x \le 1$, $4x^3 \ge 0$, so $f'(x) \ge 4 > 0$ on $[0, 1]$.

$f$ is strictly increasing on $[0, 1]$, so exactly one root there. (There is a second root near $-1.66$, outside $[0, 1]$: uniqueness is only claimed where $f' > 0$.)

$f$ is a polynomial: continuous on $[0, 2]$ and differentiable on $(0, 2)$.

$f(2) - f(0) = f'(c)(2 - 0)$, with $f(2) = 6$, $f(0) = 0$: average slope $\dfrac{6}{2} = 3$.

$f'(x) = 3x^2 - 1$.

$3c^2 - 1 = 3 \Rightarrow c^2 = \tfrac{4}{3} \Rightarrow c = \pm\tfrac{2}{\sqrt{3}}$. Keep $c = \tfrac{2}{\sqrt{3}} \approx 1.15 \in (0, 2)$.

$\sqrt{x}$ is continuous on $[1, 9]$ and differentiable for $x > 0$, so on $(1, 9)$.

$f(9) - f(1) = 3 - 1 = 2 = f'(c)(9 - 1)$: average slope $\tfrac{1}{4}$.

$f'(x) = \dfrac{1}{2\sqrt{x}}$.

$\dfrac{1}{2\sqrt{c}} = \dfrac{1}{4} \Rightarrow \sqrt{c} = 2 \Rightarrow c = 4 \in (1, 9)$.

Differentiable everywhere, so continuous on $[1, 4]$ and differentiable on $(1, 4)$.

$f(4) - f(1) = f'(c)(4 - 1)$, so $f(4) = 4 + 3f'(c)$ for some $c \in (1, 4)$.

Given: $2 \le f'(c) \le 5$.

$4 + 3(2) \le f(4) \le 4 + 3(5)$, so $10 \le f(4) \le 19$.

$f(x) = \sqrt{x}$ is continuous on $[100, 101]$ and differentiable on $(100, 101)$.

$\sqrt{101} - \sqrt{100} = f'(c)(101 - 100) = f'(c)$ for some $c \in (100, 101)$.

$f'(c) = \dfrac{1}{2\sqrt{c}}$, and $c > 100$ gives $\sqrt{c} > 10$, so $f'(c) < \dfrac{1}{20}$.

$\sqrt{101} - 10 < \dfrac{1}{20} = 0.05$, so $\sqrt{101} < 10.05$.

$[-2, 2]$ is closed and bounded.

The only problem point is $x = 3$, which is not in $[-2, 2]$, so $f$ is continuous there.

By the EVT, $f$ has a global max and a global min on $[-2, 2]$.

Not needed: the EVT already answered.

$[0, 4]$ is closed and bounded.

$x = 3$ is inside $[0, 4]$ and $f$ is undefined there: not continuous.

The EVT does not apply.

$\displaystyle\lim_{x\to 3^-} f(x) = -\infty$ and $\displaystyle\lim_{x\to 3^+} f(x) = +\infty$, so $f$ takes arbitrarily large and arbitrarily negative values: no global max, no global min.

$[0, \infty)$ is closed but not bounded.

$f$ is continuous everywhere.

The EVT does not apply directly.

Min: $f(x) \ge 0 = f(0)$, so $f(0) = 0$ is the global min. Max: $f(1) = \tfrac{1}{e} > 0$ and $\displaystyle\lim_{x\to\infty} x^2 e^{-x} = 0$, so $f(x) < \tfrac{1}{e}$ past some $R$. On $[0, R]$ the EVT gives a max $\ge \tfrac{1}{e}$, which beats everything past $R$: a global max.

$(0, 1]$ is not closed: $0$ is left out.

$f$ is continuous on $(0, 1]$.

The EVT does not apply.

$\displaystyle\lim_{x\to 0^+} \dfrac{1}{x} = \infty$: no global max. $f$ is decreasing, so its smallest value is $f(1) = 1$: a global min.