What are the local extrema of #f(x)= x^3 - 9x^2 + 19x - 3 #?

Applying the quadratic formula:

To test for local maximum or minimum:

graph{ x^3-9x^2+19x-3 [-22.99, 22.65, -10.94, 11.87]}

To find the local extrema of $f(x) = x^3 - 9x^2 + 19x - 3$, we first take the derivative of the function and set it equal to zero to find critical points. Then, we analyze the second derivative to determine the nature of these critical points.

$f'(x) = 3x^2 - 18x + 19$

Setting $f'(x)$ equal to zero:

$3x^2 - 18x + 19 = 0$

Using the quadratic formula:

$x = \frac{{18 \pm \sqrt{{(-18)^2 - 4(3)(19)}}}}{{2(3)}}$

$x = \frac{{18 \pm \sqrt{{324 - 228}}}}{{6}}$

$x = \frac{{18 \pm \sqrt{{96}}}}{{6}}$

$x = \frac{{18 \pm 4\sqrt{6}}}{{6}}$

$x = \frac{{9 \pm 2\sqrt{6}}}{{3}}$

So, the critical points are $x = \frac{{9 + 2\sqrt{6}}}{{3}}$ and $x = \frac{{9 - 2\sqrt{6}}}{{3}}$.

Now, we find the second derivative:

$f''(x) = 6x - 18$

Evaluate $f''(x)$ at the critical points:

For $x = \frac{{9 + 2\sqrt{6}}}{{3}}$, $f''\left(\frac{{9 + 2\sqrt{6}}}{{3}}\right) > 0$, so it's a local minimum.

For $x = \frac{{9 - 2\sqrt{6}}}{{3}}$, $f''\left(\frac{{9 - 2\sqrt{6}}}{{3}}\right) < 0$, so it's a local maximum.

Therefore, the local extrema of the function are:

Local minimum at $x = \frac{{9 + 2\sqrt{6}}}{{3}}$

Local maximum at $x = \frac{{9 - 2\sqrt{6}}}{{3}}$