Answer :
w has relative maximum(s) at x = -1 and x = 5(4)
w has relative minimum(s) at x = 0.
The given function is, [tex]w(x)=12x^5-60x^4-100x^3+4[/tex]
Differentiating the function w(x)
We get, [tex]w'(x) = 60x^4 - 240x^3- 300x^2[/tex]
At any point x, w'(x) represents the slope of the tangent to the curve at point x.
(1) All intervals on which w is increasing
For w to be increasing w'(x) > 0
For w to be decreasing w'(x) < 0
For w to have a relative maximum w'(x) = 0
[tex]w'(x) = 60x^4 - 240x^3- 300x^2[/tex]
=> [tex]60x^2(x^2- 4x - 5)[/tex]
=> [tex]60x^2(x - 5)(x + 1)[/tex]
Therefore, w is increasing on two intervals which are x ∈ (-∞,-1) U (0,5) (2)
All intervals on which w is decreasing
w is decreasing where w'(x) < 0
[tex]w'(x) = 60x^4 - 240x^3 - 300x^2[/tex]
=> [tex]60x^2(x^2- 4x - 5)[/tex]
=> [tex]60x^2(x - 5)(x + 1)[/tex]
Therefore, w is decreasing on two intervals which are x ∈ (-1,0) U (5,∞)(3)
The value(s) of x at which w has a relative maximum
For w to have a relative maximum w'(x) = 0
[tex]w'(x) = 60x^4 - 240x^3 - 300x^2[/tex]
=> [tex]60x^2(x^2- 4x - 5)[/tex]
=> [tex]60x^2(x - 5)(x + 1)[/tex]
x = -1 and x = 5
Therefore, w has relative maximum(s) at x = -1 and x = 5(4)
The value(s) of x at which w has a relative minimum
For w to have a relative minimum w'(x) = 0
[tex]w'(x) = 60x^4 - 240x³^3 - 300x^2[/tex]
=> [tex]60x^2(x^2 - 4x - 5)[/tex]
=> [tex]60x^2 (x - 5)(x + 1) x = 0[/tex]
Therefore, w has relative minimum(s) at x = 0.
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