Question

find the exact location of the relative and absolute extrema of the function (g(x)=-2x^{3}+24x + 3) on the interval (-5,5). (enter dne for \does not exist\ if there are none.)
(g(x)) has an absolute minimum at ((x,y)=)
(g(x)) has an absolute maximum at ((x,y)=)
(g(x)) has a relative minimum at ((x,y)=)
(g(x)) has a relative maximum at ((x,y)=)

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 \((5,-127)\)