Find critical numbers of f(x)

Differentiate: $f'(x) = 3x^2 - 6x - 9$. This is a polynomial, so it is defined for all $x$, meaning critical numbers only come from setting $f'(x) = 0$.

Factor: $3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$. Setting this equal to zero gives $x = 3$ and $x = -1$.

The critical numbers $x = -1$ and $x = 3$ split the real line into three intervals: $(-\infty, -1)$, $(-1, 3)$, and $(3, \infty)$. Testing a point in each, such as $x=-2, 0, 4$, in $f'(x) = 3(x-3)(x+1)$ determines the sign of $f'$ there.

At $x=-2$: $f'(-2) = 3(-5)(-1) = 15 > 0$. At $x=0$: $f'(0) = 3(-3)(1) = -9 < 0$. At $x=4$: $f'(4) = 3(1)(5) = 15 > 0$. So $f$ is increasing where $f'>0$ and decreasing where $f'<0$.

At $x=-1$, $f'$ changes from positive (on $(-\infty,-1)$) to negative (on $(-1,3)$), so by the First Derivative Test $f$ has a local maximum at $x=-1$. Compute $f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$.

At $x=3$, $f'$ changes from negative (on $(-1,3)$) to positive (on $(3,\infty)$), so $f$ has a local minimum at $x=3$. Compute $f(3) = 27 - 27 - 27 + 5 = -22$.

The critical number x equals negative one gives a local maximum value of ten, while x equals three gives a local minimum value of negative twenty-two. Comparing these two values, ten is larger than negative twenty-two. So the critical number x equals negative one produces the larger function value.