Answer:
Critical points: [tex]x_{1} = 0[/tex] and [tex]x_{2} = 12[/tex]
Step-by-step explanation:
First, the function, which is a division of two functions, has to be derived:
[tex]f'(x) = \frac{f(x)\cdot g'(x)-f'(x)\cdot g(x)}{[g(x)]^{2}}[/tex], where [tex]f(x) = x^{3}[/tex] and [tex]g(x) = 5\cdot (8-x)[/tex].
The derivatives of each function are, respectively:
[tex]f'(x) = 3\cdot x^{2}[/tex]
[tex]g'(x) = -5[/tex]
All components are replaced and expressed is simplified afterwards:
[tex]f'(x) = \frac{(x^{3})\cdot (-5)-(3\cdot x^{2})\cdot [5\cdot (8-x)]}{25}[/tex]
[tex]f'(x) = \frac{-5\cdot x^{3}-120\cdot x^{2}+15\cdot x^{3}}{25}[/tex]
[tex]f'(x) = \frac{10\cdot x^{3}-120\cdot x^{2}}{25}[/tex]
[tex]f'(x) = \frac{2}{5}\cdot x^{3} - \frac{24}{5}\cdot x^{2}[/tex]
[tex]f'(x) = \frac{2}{5}\cdot x^{2}\cdot(x-12)[/tex]
Let [tex]f'(x) = 0[/tex] be equalized to zero and critical numbers are found by the First Derivative Test:
[tex]\frac{2}{5}\cdot x^{2}\cdot (x-12) = 0[/tex]
The critical points are:
[tex]x_{1} = 0[/tex] and [tex]x_{2} = 12[/tex]
