The function f given by f(x)=3x5−4x3−3x has a relative maximum at x?

The function f given by has a relative maximum at

x = -1

The given function is:

Find the derivative of the function:

Find the critical points by equation f'(x) to zero

By solving the equation above, the critical points are:

x = -1,  x =  1

Find the second derivative of f(x)

Since f''(-1) < 0, x = -1 is a maximum point

Since f''(1) > 0, x = 1 is a minimum point

Therefore, the function has a relative maximum at x = -1

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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:

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

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