Answer:
a) [tex]x = -3[/tex], b) [tex]x = 0[/tex], [tex]x = -6[/tex], c) [tex]x = 0[/tex], d) [tex]x = -6[/tex]
Step-by-step explanation:
a) Let derive the function:
[tex]f'(x) = \frac{10\cdot x \cdot (x+3)-5\cdot x^{2}}{25\cdot (x+3)^{2}}[/tex]
[tex]f'(x)[/tex] is undefined when denominator equates to zero. The critical point is:
[tex]x = -3[/tex]
b) [tex]f'(x) = 0[/tex] when numerator equates to zero. That is:
[tex]10\cdot x \cdot (x+3) - 5\cdot x^{2} = 0[/tex]
[tex]10\cdot x^{2}+30\cdot x -5\cdot x^{2} = 0[/tex]
[tex]5\cdot x^{2} + 30\cdot x = 0[/tex]
[tex]5\cdot x \cdot (x+6) = 0[/tex]
This equation shows two critical points:
[tex]x = 0[/tex], [tex]x = -6[/tex]
c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that [tex]x = 0[/tex] corresponds to a local maximum. (See Attachment)
d) By plotting the function, it is evident that [tex]x = -6[/tex] corresponds to a local minimum. (See Attachment)