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Answer :
To determine the relative maximum or minimum of the function [tex]f(x) = 12x^5 - 45x^4 + 40x^3 + 5[/tex], we need to find the critical points and use the first derivative test.
Find the first derivative:
The first derivative, [tex]f'(x)[/tex], will help us find the critical points.
[tex]f'(x) = \frac{d}{dx}(12x^5 - 45x^4 + 40x^3 + 5) = 60x^4 - 180x^3 + 120x^2[/tex]
Set the first derivative equal to zero to find critical points:
Set [tex]f'(x) = 0[/tex]:
[tex]60x^4 - 180x^3 + 120x^2 = 0[/tex]
Factor out the greatest common factor:
[tex]60x^2(x^2 - 3x + 2) = 0[/tex]
Solve for [tex]x[/tex]:
- [tex]60x^2 = 0[/tex] gives [tex]x = 0[/tex]
- [tex]x^2 - 3x + 2 = 0[/tex] factors to [tex](x-1)(x-2) = 0[/tex]
This gives the solutions [tex]x = 1[/tex] and [tex]x = 2[/tex].
So, the critical points are [tex]x = 0, 1, \text{ and } 2[/tex].
Determine the nature of these critical points using the first derivative test:
We'll test the sign of [tex]f'(x)[/tex] around each critical point:
For [tex]x = 0[/tex]:
- Choose a test point less than 0, e.g., [tex]x = -1[/tex], [tex]f'(-1) > 0[/tex]
- Choose a test point between 0 and 1, e.g., [tex]x = 0.5[/tex], [tex]f'(0.5) < 0[/tex]
- This indicates a relative maximum at [tex]x = 0[/tex].
For [tex]x = 1[/tex]:
- Choose a test point between 0 and 1, e.g., [tex]x = 0.5[/tex], [tex]f'(0.5) < 0[/tex]
- Choose a test point between 1 and 2, e.g., [tex]x = 1.5[/tex], [tex]f'(1.5) > 0[/tex]
- This indicates a relative minimum at [tex]x = 1[/tex].
For [tex]x = 2[/tex]:
- Choose a test point greater than 2, e.g., [tex]x = 3[/tex], [tex]f'(3) > 0[/tex]
- This indicates a relative maximum at [tex]x = 2[/tex].
Conclusion:
- There is a relative maximum at [tex]x = 0[/tex] and [tex]x = 2[/tex].
- There is a relative minimum at [tex]x = 1[/tex].
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