find the exact location of the relative and absolute extrema of the function (g(x)=-2x^{3}+24x + 3) on the…

Explanation:

Step1: Find the derivative of (g(x))

First, find (g'(x)) for (g(x)=- 2x^{3}+24x + 3). Using the power - rule ((x^n)'=nx^{n - 1}), we have (g'(x)=-6x^{2}+24).

Step2: Set the derivative equal to zero and solve for (x)

Set (g'(x) = 0), so (-6x^{2}+24 = 0). Rearrange to get (6x^{2}=24), then (x^{2}=4), and (x=\pm2). These are the critical points.

Step3: Use the second - derivative test to find relative extrema

Find the second - derivative (g''(x)). Differentiating (g'(x)=-6x^{2}+24) gives (g''(x)=-12x). When (x = 2), (g''(2)=-12\times2=-24<0), so (g(x)) has a relative maximum at (x = 2). Calculate (y): (g(2)=-2(2)^{3}+24(2)+3=-16 + 48+3=35). So the relative maximum is at ((2,35)). When (x=-2), (g''(-2)=-12\times(-2)=24>0), so (g(x)) has a relative minimum at (x=-2). Calculate (y): (g(-2)=-2(-2)^{3}+24(-2)+3 = 16-48 + 3=-29). So the relative minimum is at ((-2,-29)).

Step4: Evaluate the function at the endpoints of the interval ([-5,5])

Evaluate (g(x)) at (x=-5): (g(-5)=-2(-5)^{3}+24(-5)+3=250-120 + 3=133). Evaluate (g(x)) at (x = 5): (g(5)=-2(5)^{3}+24(5)+3=-250+120 + 3=-127).

Step5: Determine the absolute extrema

Compare the values of (g(x)) at the critical points and endpoints. (g(-5)=133), (g(-2)=-29), (g(2)=35), (g(5)=-127). The absolute maximum occurs at (x=-5) and (y = 133), so the absolute maximum is at ((-5,133)). The absolute minimum occurs at (x = 5) and (y=-127), so the absolute minimum is at ((5,-127)).

Answer:

Relative maximum at ((2,35)) Relative minimum at ((-2,-29)) Absolute maximum at ((-5,133)) Absolute minimum at ((5,-127))