Solutions

1. $f'(x) = 3x^2 - 2x - 8$

The critical points are

$$\begin{align*} f'(x) = 0 &\implies 3x^2 - 2x - 8 = 0 \\ &\implies 3x^2 - 6x + 4x - 8 = 0 \\ &\implies 3x(x - 2) + 4(x - 2) = 0 \\ &\implies (x - 2)(3x + 4) = 0 \\ &\implies x = 2 \;\vee\; x = -\frac{4}{3}. \end{align*}$$

And the critical values are

$$f(2) = (2)^3 - (2)^2 - 8(2) + 1 = -11$$

and

$$\begin{align*} f\left(-\frac{4}{3}\right) &= \left(-\frac{4}{3}\right)^3 - \left(-\frac{4}{3}\right)^2 - 8\left(-\frac{4}{3}\right) + 1 \\ &= -\frac{64}{27} - \frac{16}{9} + \frac{32}{3} + 1 \\ &= \frac{203}{27}. \end{align*}$$

We also have $f(-2) = 5$. Hence, $f\left(-\frac{4}{3}\right)$ is the maximum value and $f(2)$ is the minimum value.

2. $f'(x) = 5x^4 + 1$

Note that there are no critical points since

$$\begin{align*} f'(x) = 0 &\implies 5x^4 + 1 = 0 \\ &\implies x^4 = -\frac{1}{5}. \end{align*}$$

The values at the boundaries are

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

Thus $f(-1)$ is the minimum value and $f(1)$ is the maximum value.

3. $f'(x) = 12x^3 - 24x^2 + 12x$

The critical points are

$$\begin{align*} f'(x) = 0 &\implies 12x(x^2 - 2x + 1) = 0 \\ &\implies 12x(x - 1)^2 = 0 \\ &\implies x = 0 \;\vee\; x = 1. \end{align*}$$

The critical values are $f(0) = 0$ and $f(1) = 1$. We also have $f\left(-\frac{1}{2}\right) = \frac{43}{16}$ and $f\left(\frac{1}{2}\right) = \frac{11}{16}$. Hence, the minimum value is $f(0) = 0$ and the maximum value is $f\left(-\frac{1}{2}\right) = \frac{43}{16}$.

$$f'(x) = -\frac{5x^4 + 1}{(x^5 + x + 1)^2}.$$

There are no critical points but $f'$ is negative everywhere on its domain. We have $f\left(-\frac{1}{2}\right) = \frac{32}{17}$ and $f(1) = \frac{1}{3}$. Hence, the minimum value is $f(1)$ and the maximum value is $f\left(-\frac{1}{2}\right)$.

$$\begin{align*} f'(x) &= \frac{x^2 + 1 - (x + 1)(2x)}{(x^2 + 1)^2} \\ &= \frac{-x^2 - 2x + 1}{(x^2 + 1)^2} \\ &= \frac{2 - (x + 1)^2}{(x^2 + 1)^2}. \end{align*}$$

The critical points are

$$\begin{align*} f'(x) = 0 &\implies 2 = (x + 1)^2 \\ &\implies \pm\sqrt{2} = x + 1 \\ &\implies x = -1 \pm \sqrt{2}. \end{align*}$$

We have the critical value

$$f(-1 + \sqrt{2}) = \frac{1}{2(\sqrt{2} - 1)}.$$

Also, $f(-1) = 0$ and $f\left(\frac{1}{2}\right) = \frac{6}{5}$. Thus, the minimum is $f(-1)$ and the maximum value is $f(-1 + \sqrt{2})$.

$$\begin{align*} f'(x) &= \frac{x^2 - 1 - x(2x)}{x^2 - 1} \\ &= -\frac{x^2 + 1}{x^2 - 1}. \end{align*}$$

There are no real critical points. The interval $[0, 5]$ includes the point $x = 1$ where $f$ is undefined. We have

$$\lim_{x \to 1^-} \frac{x}{x^2 - 1} = -\infty \qquad\text{and}\qquad \lim_{x \to 1^+} \frac{x}{x^2 - 1} = \infty.$$

Hence, there are no minimum or maximum values.