The given function f(x) is defined by:
[tex]f(x)=x^3+5x^2-4x-4[/tex]The critical points is found where f'(x)=0
[tex]f^{\prime}(x)=3x^2+10x-4[/tex]
When f'(x) = 0
[tex]3x^2+10x-4=0[/tex]Solve the equation graphically:
Therefore, the critical points are at x = -3.694 and 0.361
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