The function is given to be:
[tex]f\left(x\right)=12x^5+15x^4-80x^3-5[/tex]
The graph of the function is shown below:
The critical points are points where the function is defined and its derivative is zero or undefined.
The derivative is calculated to be:
[tex]f^{\prime}(x)=60x^4+60x^3-240x^2[/tex]
The critical points are at f'(x) = 0:
[tex]\begin{gathered} 60x^4+60x^3-240x^2=0 \\ Simplifying \\ x^4+x^3-4x^2=0 \end{gathered}[/tex]
Factoring:
[tex]x^2(x^2+x-4)=0[/tex]
Therefore, applying the zero factor principle:
[tex]\begin{gathered} x^2=0 \\ \therefore \\ x=0 \end{gathered}[/tex]
or
[tex]\begin{gathered} x^2+x-4=0 \\ Using\text{ }the\text{ }Quadratic\text{ }Formula \\ x=1.56155,x=-2.56155 \end{gathered}[/tex]
Therefore, the critical numbers are given below with their descriptions:
[tex]\begin{gathered} x=-2.56155,\text{ Local Maximum} \\ x=0,\text{ Neither a maximum or a minimum} \\ x=1.56155,\text{ Local Minimum} \end{gathered}[/tex]
