How do you find the exact relative maximum and minimum of the polynomial function of $g (x) = x^{3} - 3 x^{2} - 9 x + 1$ $g(x) = x^3 - 3x^2 - 9x +1$ ?

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Answer 1 · 2016-03-16T17:24:09

$P_{max} \to (x, y) \to (- 1, 6)$ $P_("max") ->(x,y) ->(-1,6)$
$P_{min} \to (x, y) \to (3, - 26)$ $P_("min")->(x,y)->(3,-26)$

$\frac{d y}{d x} = 3 x^{2} - 6 x - 9 = 0$ $(dy)/(dx)=3x^2-6x-9 = 0$

Divide throughout by 3 giving

$x^{2} - 2 x - 3 = 0$ $x^2-2x-3=0$

$(x + 1) (x - 3) = 0$ $(x+1)(x-3)=0$

$x = - 1 or + 3$ $x=-1" or "+3$
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$\frac{d^{2} y}{{d x}^{2}} = 6 x - 6$ $(d^2y)/(dx^2)=6x-6$

At $x = - 1; \frac{d^{2} y}{{d x}^{2}} < 0 \Rightarrow maximum$ $x=-1 ; (d^2y)/(dx^2)<0 => "maximum"$

At $x = + 3; \frac{d^{2} y}{{d x}^{2}} > 0 \Rightarrow minimum$ $x=+3 ; (d^2y)/(dx^2)>0 => "minimum"$

By substitution you find

$P_{max} \to (x, y) \to (- 1, 6)$ $P_("max") ->(x,y) ->(-1,6)$
$P_{min} \to (x, y) \to (3, - 26)$ $P_("min")->(x,y)->(3,-26)$

Tony B

How do you find the exact relative maximum and minimum of the polynomial function of g(x)=x3−3x2−9x+1g(x) = x^3 - 3x^2 - 9x +1?