Respuesta :
Answer:
We have the function:
f(x) = 3*x^5 - 4*x^3 - 3*x
To find the maximum/minimums, we first need to look at the roots of the first derivative of f(x), this is:
f´(x) = 5*3*x^4 - 3*4*x^2 - 3
f´(x) = 15*x^4 - 12*x^2 - 3
We can do the change of variable:
u = x^2
Then we can rewrite:
f'(u) = 15*u^2 - 12*u - 3
This is a quadratic equation, the solutions will be obtained by the Bhaskara´s equation:
So the solutions are:
[tex]u = \frac{+12 +- \sqrt{(-12)^2 - 4*(-3)*15} }{2*15} = \frac{12 +- 18}{30}[/tex]
Then the two solutions are:
u = (12 + 18)/30 = 1
this leads to:
x^2 = 1, then we can have x = ± 1
the other solution is:
u = (12 - 18)/30 = -6/30 = -(1/5)
and x^2 = -(1/5), then:
x = ±√(1/5)*i
So those are the possible values for the minimum/maximums.
Now to see it, we need to look at the second derivative ad see iff:
f´´(x) > 0 then x is a minimum
f''(x) < 0, then x is a maximum.
We have that:
f´´(x) = 4*15*x^3 - 2*12*x
f´´(x) = 60*x^3 - 24*x
Notice that the two complex roots will lead to complex values of f(x), so these can be ignored here, then we have to try only with x = ± 1.
f´´(1) = 60*1^3 - 24*1 = 36
then x = 1 is a minimum.
f´´(-1) = 60*(-1)^3 -24*-1 = -36
then x = -1 is a maximum.
Then we can conclude that the relative maximum is at x = -1
The function f given by [tex]f(x) = 3x^5-4x^3-3x[/tex] has a relative maximum at
x = -1
The given function is:
[tex]f(x) = 3x^5-4x^3-3x[/tex]
Find the derivative of the function:
[tex]f'(x)=15x^4-12x^2-3[/tex]
Find the critical points by equation f'(x) to zero
[tex]15x^4-12x^2-3=0\\\\5x^4-4x^2-1=0[/tex]
By solving the equation above, the critical points are:
x = -1, x = 1
Find the second derivative of f(x)
[tex]f''(x)=60x^3-24x[/tex]
[tex]f''(-1)=60(-1)^3-24(-1)\\\\f''(-1)=-36[/tex]
Since f''(-1) < 0, x = -1 is a maximum point
[tex]f''(1)=60(1)^3-24(1)\\\\f''(1)=36[/tex]
Since f''(1) > 0, x = 1 is a minimum point
Therefore, the function [tex]f(x) = 3x^5-4x^3-3x[/tex] has a relative maximum at x = -1
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