Given the function f(x) defined as:
[tex]f(x)=x+\frac{4}{x^2}[/tex]
Taking the derivative of the function:
[tex]f^{\prime}(x)=1-\frac{8}{x^3}[/tex]
Now, we calculate the critical points using the equation:
[tex]\begin{gathered} f^{\prime}(x)=0 \\ 1-\frac{8}{x^3}=0 \\ 1=\frac{8}{x^3} \\ x^3=8 \\ x=2 \end{gathered}[/tex]
Now, we explore the intervals of increasing and decreasing for f(x) using the critical point, taking into account the discontinuity in x = 0:
[tex]\begin{gathered} x>2\colon f^{\prime}(x)>0\text{ (Increasing)} \\ 00\text{ (Increasing)} \end{gathered}[/tex]
The intervals are:
[tex]\begin{gathered} \text{Increasing}\colon(-\infty,0)\cup(2,\infty) \\ \text{Decreasing}\colon(0,2) \end{gathered}[/tex]
The function has a local minimum at x = 2, because f''(2) > 0. There is no local maximum.
