Certainly! Here’s the transcription for the educational content:---### Calculus Problem Set: Absolute Extrema**Problem 2:**(a) **Objective**: Find the **absolute maximum** and **absolute minimum** values of the function:\[ g(x) = (x^2 - 1)^{\frac{1}{3}} \]on the closed interval \([-2, 3]\).**Question**: What test/method is **most appropriate** for this exercise and **why**? *Hint: It’s **not** the first derivative test.*(b) **Task**: Solve the problem and **show all steps**.---### DiscussionThis problem involves finding the extreme values of a function on a closed interval, a common task in calculus. The function \( g(x) \) is defined on the interval \([-2, 3]\), and you are guided to use a method other than the first derivative test to find the absolute extrema. ---

Given g(x) = (x2-1)13, [-2,3] g'(x)=2x3x2-123 g''(x)=-29×x2+3x2-153 Note that x=-1,0,1 are critical points because the derivative is 0 for x=0 and derivative does not exist for points x=-1,1, but x=-1,1 are in the domain of the given function Since both the first and second derivative does not exist on the critical points x=-1,1 therefore these are not suitable to find the local or absolute minima or maxima of the function (a) To solve this problem graphical method is most appropriate to solve this exercise (b) The graph of the function g(x) = (x2-1)13 on the domain [-2,3] is given as Note that from the derivative g'(x)=2x3x2-123 it can be obtained that x=-1.0,2 are critical points From the graph note that x=-1,1 does not correspond to local minima or maxima but x=0 corresponds to a local minimum. Since the function is continuous on the interval [-2,3] then the minima or maxima may occur at the endpoints also. at x=3, g(x) = 2 and at x=-2, g(x) = 1.442 for the critical point x=0, g(0) = -1 therefore the function have absolute minima at x = 0 and the minimum value is -1 and function have maximum value at the boundary point x = 3 and the maximum value is 2.