Question

what is the maximum number of relative extrema contained in the graph of this function?
$f(x)=3x^{5}-x^{3}+4x - 2$

Explanation:

Step1: Recall the degree of the function

The given function \( f(x)=3x^{5}-x^{3}+4x - 2 \) is a polynomial function. The degree \( n\) of a polynomial \( y = a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{1}x + a_{0}\) is \(n = 5\) (the highest power of \(x\)).

Step2: Use the relationship between degree and number of relative extrema

For a polynomial function \(y=f(x)\) of degree \(n\), the number of critical points (found by \(f^{\prime}(x)=0\)) is at most \(n - 1\). The first - derivative of \(f(x)\) is \(f^{\prime}(x)=15x^{4}-3x^{2}+4\). The degree of \(f^{\prime}(x)\) is \(n-1 = 4\). The number of relative extrema (which occur at critical points where the concavity changes, i.e., \(f^{\prime\prime}(x)
eq0\) at those critical points) is at most the number of critical points. A polynomial of degree \(m\) has at most \(m\) real roots. Since \(f^{\prime}(x)\) is a polynomial of degree \(4\), the equation \(f^{\prime}(x) = 0\) has at most \(4\) real roots.

Answer:

\(4\)