Question

EXAMPLE 7 Find the critical numbers of the function. f(x) = x3/5(8 − x). SOLUTION The Product Rule gives the following. f '(x) = x3/5 Correct: Your answer is correct. + (8 − x) Correct: Your answer is correct. = Correct: Your answer is correct. + 3(8 − x) Correct: Your answer is correct. = Incorrect: Your answer is incorrect. + 3(8 − x) Correct: Your answer is correct. =

Answer 1

Answer:

Critical points: $x_{1} = 0$ and $x_{2} = 12$

Step-by-step explanation:

First, the function, which is a division of two functions, has to be derived:

$f'(x) = \frac{f(x)\cdot g'(x)-f'(x)\cdot g(x)}{[g(x)]^{2}}$, where $f(x) = x^{3}$ and $g(x) = 5\cdot (8-x)$.

The derivatives of each function are, respectively:

$f'(x) = 3\cdot x^{2}$

$g'(x) = -5$

All components are replaced and expressed is simplified afterwards:

$f'(x) = \frac{(x^{3})\cdot (-5)-(3\cdot x^{2})\cdot [5\cdot (8-x)]}{25}$

$f'(x) = \frac{-5\cdot x^{3}-120\cdot x^{2}+15\cdot x^{3}}{25}$

$f'(x) = \frac{10\cdot x^{3}-120\cdot x^{2}}{25}$

$f'(x) = \frac{2}{5}\cdot x^{3} - \frac{24}{5}\cdot x^{2}$

$f'(x) = \frac{2}{5}\cdot x^{2}\cdot(x-12)$

Let $f'(x) = 0$ be equalized to zero and critical numbers are found by the First Derivative Test:

$\frac{2}{5}\cdot x^{2}\cdot (x-12) = 0$

The critical points are:

$x_{1} = 0$ and $x_{2} = 12$