Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 60

a.) Given the function $f(x) = x^4 - 3x^3 + 3x^2 - x, \quad 0 \leq x \leq 2$, use a graph to estimate the absolute maximum and minimum values.
b.) Use calculus to find the exact maximum and minimum values.

a.)



Based from the graph, the absolute maximum is approximately at $f(2) \approx 2$. While the absolute minimum is approximately at $f(0.2) \approx -0.15$

b.) To find the exact value, we take the derivative of the function
$\displaystyle f'(x) = \frac{d}{dx} (x^4) - 3 \frac{d}{dx} (x^3) + 3 \frac{d}{dx} (x^2) - \frac{d}{dx} (x)$
$\displaystyle f'(x) = 4x^3 - 9x^2 + 6x -1$

When $f'(x) = 0$
$ 0 = 4x^3 - 9x^2 + 6x - 1$

By factoring we get,
$ (4x - 1)(x-1)^2 = 0$

So,
$\displaystyle x = \frac{1}{4}$ and $x = 1$

When $\displaystyle \frac{1}{4}$,
$\displaystyle f \left( \frac{1}{4} \right) = \left( \frac{1}{4} \right)^4 - 3 \left( \frac{1}{4} \right)^3 + 3 \left( \frac{1}{4} \right)^2 - \left( \frac{1}{4} \right)$
$\displaystyle f \left( \frac{1}{4} \right) = -0.1055$

When $x = 1$,
$f(1) = (1)^4 - 3 (1)^3 + 3 (1)^2 - (1)$
$f(1) = 0$

Also, we evaluate $f(x)$ with the end points of $x = 0 $ and $x = 2$
So when $x = 0$,
$f(0) = (0)^4 - 3 (0)^3 + 3 (0)^2 - (0)$
$f(0) = 0$

when $x = 2$,
$f(2) = (2)^4 - 3(2)^3 + 3 (2)^2 - (2)$
$f(2) = 2$

Therefore, the exact value of the absolute maximum and minimum values are $f(2) = 2$ and $\displaystyle f \left( \frac{1}{4}\right) = -0.1055$ respectively.