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

To find the maximum and minimum values of the function, we first need to calculate 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, we find:

$x = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}$

This gives us the stationary points:

$x = 3 \quad \text{or} \quad x = \frac{1}{3}$

Now, we evaluate f at the endpoints of the interval and at the stationary points:

• $f(0) = 5$
• $f(\frac{1}{3}) = \frac{1}{27} \times 1 + 5 - \frac{5}{9} + 3$ (calculate this to get approximately 1.48)
• $f(3) = 27 - 5(9) + 3(3) + 5 = -4$
• $f(5) = 125 - 125 + 15 + 5 = 20$

Thus, the maximum value of f on $A$ is 20 and the minimum value is -4.