A is the closed interval [0, 5] - Leaving Cert Mathematics - Question 5 - 2012

To find the absolute maximum and minimum values of the function, we first compute the derivative of f:

$$f'(x) = 3x^2 - 10x + 3.$$

Next, we set the derivative equal to zero to find the stationary points:

$$3x^2 - 10x + 3 = 0.$$

Using the quadratic formula:

x = rac{-b oldsymbol{ extpm} ext{sqrt}{b^2 - 4ac}}{2a} = rac{10 oldsymbol{ extpm} ext{sqrt}{(-10)^2 - 4(3)(3)}}{2(3)}

This yields:

x = rac{10 oldsymbol{ extpm} ext{sqrt}{100 - 36}}{6} = rac{10 oldsymbol{ extpm} 8}{6} ightarrow x = 3 ext{ or } x = rac{1}{3}.

Both stationary points lie within the interval A = [0, 5]. Thus, we evaluate the function at the endpoints and at the stationary points:

1. At x = 0: $f(0) = 5$

2. At x = \frac{1}{3}: $f \left(\frac{1}{3}\right) = \frac{1}{27} - \left(\frac{5}{9}\right) + 1 + 5 = \frac{148}{27}$

3. At x = 3: $f(3) = 27 - 45 + 9 + 5 = -4$

4. At x = 5: $f(5) = 125 - 125 + 15 + 5 = 20$

The maximum value of f is 20, and the minimum value of f is -4.