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You need to remember that you may find the maximum and minimum points of the function solving the equation `f'(x) = 0` such that:
`f'(x) = (2x^6 + 3x^5 + 3x^3 - 2x^2)'`
`f'(x) = (12x^5 + 15x^4 + 9x^2 - 4x)`
You need to solve the equation `f'(x) = 0` such that:
`12x^5 + 15x^4 + 9x^2 - 4x = 0`
You need to factor out x such that:
`x(12x^4 + 15x^3 + 9x - 4) = 0`
`x = 0`
`12x^4 + 15x^3 + 9x - 4 = 0 `
You need to write the factored form of polynomial `12x^4 + 15x^3 + 9x - 4` such that:
`12x^4 + 15x^3 + 9x - 4 = (x + 1.616)(x - 0.352)(0.007 + 0.766i)(0.007- 0.766i)`
Hence, the function reaches its extreme points at `x = -1.616,x = 0, x = 0.352` .
Notice that the function reches its minimum at x = -1.616.
how to get 12x^4+15x^3+9x-4=(x+1.616)(x-0.352)(0.007+0.766i)(0.007-0.766)? is it any formulae to calculate it? how about f''(x)=?
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