Answer:
The graph of this function does not include any relative extrema.
Step-by-step explanation:
The point [tex]x_{0}[/tex] is an extrema of the graph of [tex]f(x)[/tex] if and only if:
To find the extrema for the given graph, start by differentiating the function of the graph:
[tex]f^{\prime}(x) = 15\, x^{4} - 3\, x^{2} + 4[/tex].
The expression above should be [tex]0[/tex] at each extrema (if any) of [tex]f(x)[/tex]. To find these points, set [tex]f^{\prime}(x) = 0[/tex] and solve for [tex]x[/tex].
[tex]15\, \left(x^{2}\right)^{2} - 3\, \left(x^{2}\right) + 4 = 0[/tex].
Notice that this equation is equivalent to a quadratic equation with respect to [tex]x^{2}[/tex]. The quadratic determinant of this equation is [tex](-231)[/tex] (a negative value,) meaning that this equation has no real roots. In other words, [tex]f^{\prime}(x)[/tex] would be non-zero for all real value of [tex]x^{2}[/tex]- and by extension, all real values of [tex]x[/tex].
Because the first derivative [tex]f^{\prime}(x)[/tex] is non-zero for all real values of [tex]x[/tex], the graph of this function would not include any relative extrema.
