How do you graph $f (x) = x^{3} - 3 x^{2} - 9 x + 6$ $f(x)=x^3-3x^2-9x+6$ using the information given by the first derivative?

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Answer 1 · 2016-11-02T10:37:28

$f (x)$ $f(x)$ has a local maximun at $(- 1, 11)$ $(-1, 11)$ and a local minimum at $(3, - 21)$ $(3, -21)$

$f (x) = x^{3} - 3 x^{2} - 9 x + 6$ $f(x) = x^3-3x^2-9x+6$

$f'(x) = 3x^2-6x-9$

$f(x)$ will have turning points where $f'(x) =0$

$f(x)=0 -> 3x^2-6x-9 = 0$

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

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

$:. f(x)$ has turning points at $x = 3$ and $x=-1$

Now consider, $f''(x) = 6x-6$

$f''(3) = 12 > 0 -> x=3$ is a local minimum
$f''(-1) = -12 < 0 -> x=-1$ is a local maximum

These points can be seen on the graph of $f(x)$ below:

graph{x^3-3x^2-9x+6 [-43.74, 46.25, -24.88, 20.13]}

How do you graph f(x)=x^3-3x^2-9x+6f(x)=x3−3x2−9x+6f(x)=x^3-3x^2-9x+6 using the information given by the first derivative?