Find the coordinates of all relative and absolute extrema. $$g ( x ) = \frac { x ^ { 3 } } { x ^ { 2 } - 48 }$$
A)(-16, 16), (0, 0)
B) (-12, -18), (12, 18), (0, 0)
C) (0, 0), (12, 18)
D) (-12, -18), (12, 18)
E) (0, 0)

Question

Find the coordinates of all relative and absolute extrema.   $$g ( x ) = \frac { x ^ { 3 } } { x ^ { 2 } - 48 }$$ 
A)(-16, 16), (0, 0)
B) (-12, -18), (12, 18), (0, 0)
C) (0, 0), (12, 18)
D) (-12, -18), (12, 18)
E) (0, 0)

Quizplus · Accepted Answer

To find the extrema of the function $g(x) = \frac{x^3}{x^2 - 48}$, we first need to find its derivative and set it equal to zero to find critical points. The derivative is found using the quotient rule:$$g'(x) = \frac{(x^2 - 48)(3x^2) - (x^3)(2x)}{(x^2 - 48)^2} = \frac{3x^4 - 144x^2 - 2x^4}{(x^2 - 48)^2} = \frac{x^4 - 144x^2}{(x^2 - 48)^2}$$Setting $g'(x) = 0$ gives:$$x^4 - 144x^2 = 0$$$$x^2(x^2 - 144) = 0$$$$x^2 = 0, x^2 = 144$$$$x = 0, x = \pm 12$$To determine if these points are maxima, minima, or neither, we can use the second derivative test or analyze the sign changes of the first derivative around these points. However, for the purpose of this question, we recognize that these are the critical points where the function's slope changes.The function is undefined at $x^2 = 48$ or $x = \pm \sqrt{48} = \pm 4\sqrt{3}$, which are not among our critical points, so we don't have to worry about them for extrema within the domain of the function.Evaluating the function at the critical points:- At $x = 0$, $g(0) = \frac{0^3}{0^2 - 48} = 0$, so we have a point (0, 0).- At $x = 12$, $g(12) = \frac{12^3}{12^2 - 48} = \frac{1728}{144 - 48} = \frac{1728}{96} = 18$, so we have a point (12, 18).- At $x = -12$, $g(-12) = \frac{-12^3}{(-12)^2 - 48} = \frac{-1728}{144 - 48} = \frac{-1728}{96} = -18$, so we have a point (-12, -18).Therefore, the correct answer is D, which includes the points (-12, -18) and (12, 18).

Find the Coordinates of All Relative and Absolute Extrema g(x)=x3x2−48g ( x ) = \frac { x ^ { 3 } } { x ^ { 2 } - 48 }g(x)=x2−48x3​

Find the Coordinates of All Relative and Absolute Extrema $g ( x ) = \frac { x ^ { 3 } } { x ^ { 2 } - 48 }$