Question

Problem #10: [2 marks] Find the two critical values of the function f(x) = x2/5 (6 – 3x). Problem #10: separate your answers with a comma

Answer 1

$\\ 10. \\ \\ Critical \ \ points \ \ are \ \ where \ \ f'(x) = 0 \\ \\ Using \ \ chain \ \ rule \\ \\ Also, \ \ \frac{d(x^n)}{dx} = nx^{n - 1} \\ \\ f(x) = x^{2/5}*(6 - 3x) \\ \\ f'(x) = \frac{2}{5}*x^{2/5 - 1}*(6 - 3x) + x^{2/5}*(0 - 3*1) \\ \\ f'(x) = \frac{2(6 - 3x)}{5x^{3/5}} - 3x^{2/5} \\ \\ for \ \ critical \ \ points \\ \\ f'(x) = 0 = \frac{2(6 - 3x)}{5x^{3/5}} - 3x^{2/5} \\ \\ x^{2/5}*(\frac{2(6 - 3x)}{5x} - 3) = 0 \\ \\ (i) x^{2/5} = 0, \ \ So, \ \ x = 0 \\ \\ (ii) (\frac{2(6 - 3x)}{5x} - 3) = 0 \\ \\ 2(6 - 3x) = 15x \\ \\ 12 - 6x = 15x \\ \\ 21x = 12 \\ \\ x = \frac{12}{21} = \frac{4}{7} \\ \\ {\color{Red} Critical \ \ points: \ \ 0, \frac{4}{7}} \\$