i want to find the critical point of the graph i done that, but have difficulties then f'(x)=4x^3+3x^2-36x+24 then f'(x)= 0 is critical point, but i don't know how to solve f'(x)=0.........
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Originally Posted by jackopp i want to find the critical point of the graph i done that, but have difficulties then f'(x)=4x^3+3x^2-36x+24 then f'(x)= 0 is critical point, but i don't know how to solve f'(x)=0......... f'(x) = 0 has no rational roots ... you'll need to use technology to approximate them.
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