#### Answer

Absolute Maximum: $(0,5)$
Absolute Minimum: $(-3,-76)$

#### Work Step by Step

Original equation:$ f(x)=x^{3} - 6x^{2}+5$ on the interval $[-3,5]$
To find the absolute maximum and minimum, we first need to find all the critical points of the function, as the local extrema might be the absolute extrema. Critical points of a function occur when a function's derivative is undefined or the derivative is zero.
In this case, we see that the function is a polynomial, which means it is defined on all intervals from $(â\infty,\infty)$. Therefore, we need to set the derivative to zero.
Finding the derivative of the function through the power rule:
$fâ˛(x)=3x^{2}-12x$
Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when x=0,4
$3x^{2}-12x=0$
$x^{2}-4x=0$
$x(x-4)=0$
$x=0, 4$
Now we have four possible points of where a absolute minimum or maximum may occur: the critical points $x=0,4$, as well as the two endpoints of the interval, where $x=-3,5$. To find which one of these is an absolute extrema vs a local extrema, we plug these x values back into the original equation and see what we get.
$f(-3)=(-3)^{3} - 6(-3)^{2}+5=-76$
$f(0)=(0)^{3} - 6(0)^{2}+5=5$
$f(4)=(4)^{3} - 6(4)^{2}+5=-27$
$f(5)=(5)^{3} - 6(5)^{2}+5=-20$
Of the three values, we see that the minimum y value occurs when x=-3 and the maximum y value occurs when x=0.
Final Answer:
Absolute Maximum: $(0,5)$
Absolute Minimum: $(-3,-76)$