QUESTION IMAGE
Question
find the exact location of the relative and absolute extrema of the function ( g(x) = -2x^3 + 6x + 4 ) on the interval (-3, 3). (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) = ) ( , )
Step1: Find the derivative of \( g(x) \)
To find the extrema, we first find the derivative of \( g(x) = -2x^3 + 6x + 4 \). Using the power rule, the derivative \( g'(x) \) is:
\( g'(x) = -6x^2 + 6 \)
Step2: Find critical points
Set \( g'(x) = 0 \) to find critical points:
\( -6x^2 + 6 = 0 \)
Divide both sides by -6:
\( x^2 - 1 = 0 \)
Factor:
\( (x - 1)(x + 1) = 0 \)
So, the critical points are \( x = 1 \) and \( x = -1 \).
Step3: Analyze the function at critical points and endpoints
We evaluate \( g(x) \) at \( x = -3, -1, 1, 3 \):
- For \( x = -3 \):
\( g(-3) = -2(-3)^3 + 6(-3) + 4 = -2(-27) - 18 + 4 = 54 - 18 + 4 = 40 \)
- For \( x = -1 \):
\( g(-1) = -2(-1)^3 + 6(-1) + 4 = -2(-1) - 6 + 4 = 2 - 6 + 4 = 0 \)
- For \( x = 1 \):
\( g(1) = -2(1)^3 + 6(1) + 4 = -2 + 6 + 4 = 8 \)
- For \( x = 3 \):
\( g(3) = -2(3)^3 + 6(3) + 4 = -2(27) + 18 + 4 = -54 + 18 + 4 = -32 \)
Step4: Determine relative extrema
To find relative extrema, we can use the second derivative test. The second derivative \( g''(x) = -12x \).
- At \( x = -1 \): \( g''(-1) = -12(-1) = 12 > 0 \), so \( g(x) \) has a relative minimum at \( x = -1 \). The \( y \)-value is \( g(-1) = 0 \), so the point is \( (-1, 0) \).
- At \( x = 1 \): \( g''(1) = -12(1) = -12 < 0 \), so \( g(x) \) has a relative maximum at \( x = 1 \). The \( y \)-value is \( g(1) = 8 \), so the point is \( (1, 8) \).
Step5: Determine absolute extrema
From the values calculated:
- Absolute maximum: The largest value among \( g(-3) = 40 \), \( g(-1) = 0 \), \( g(1) = 8 \), \( g(3) = -32 \) is \( 40 \) at \( x = -3 \), so absolute maximum at \( (-3, 40) \).
- Absolute minimum: The smallest value is \( -32 \) at \( x = 3 \), so absolute minimum at \( (3, -32) \).
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- \( g(x) \) has an absolute minimum at \( (x,y) = (3, -32) \)
- \( g(x) \) has an absolute maximum at \( (x,y) = (-3, 40) \)
- \( g(x) \) has a relative minimum at \( (x,y) = (-1, 0) \)
- \( g(x) \) has a relative maximum at \( (x,y) = (1, 8) \)