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Given:


4x^3-18x^2-48x-3

find: critical number of point of a function.

Step-by-step explanation: critical numbers are values of a function where the function's tangent lines.


\begin{gathered} y=4x^3-18x^2-48x-3 \\ (dy)/(dx)=12x^2-36x-48=0 \\ x^2-3x-4=0 \\ x^2-(4-1)x-4=0 \\ x^2-4x+x-4=0 \\ x(x-4)+(x-4)=0 \\ (x-4)(x+1)=0 \\ x=4,-1 \end{gathered}

Final answer: there are two critical number x=4 and -1.

User Stephen Mesa
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