# Is f(x)=(x+3)(x-8)(x-3) increasing or decreasing at x=1?

Jun 21, 2016

decreasing

#### Explanation:

i expect your supposed to follow through on a full product rule etc but there are other ways to look at this.

first compute $f \left(1\right) = 4 \left(- 7\right) \left(- 2\right) = 56$

then compute $\delta f = f \left(1 + h\right) - f \left(1\right)$ where $h$ is a small displacement such that $h \setminus \to 0$

$\delta f = f \left(1 + h\right) - f \left(1\right) = \left(4 + h\right) \left(- 7 + h\right) \left(- 2 + h\right) - f \left(1\right)$

$= \left(4 + h\right) \left(14 - 9 h + {h}^{2}\right) - f \left(1\right)$

the neat thing is we only need look at constant and ${h}^{1}$ terms, ignoring higher order ${h}^{2} , {h}^{3} , \ldots$ terms, because $h \setminus \to 0$ so those higher order terms are orders of magnitude smaller.

so
$\delta f = \left(4 + h\right) \left(14 - 9 h + {h}^{2}\right) - f \left(1\right)$

$= 56 + h \left(- 36 + 14\right) - 56 + \setminus m a t h c a l \left\{O\right\} \left({h}^{2}\right)$

$= - 22 h + \setminus m a t h c a l \left\{O\right\} \left({h}^{2}\right)$

now for $h > 0$, we have $\delta f < 0$. for $h < 0$, we have $\delta f > 0$ so this function is decreasing as it passes through $x = 1$

moreover, the actual slope at $x = 1$ should be given by ${\lim}_{h \setminus \to 0} {\left[\frac{\delta f}{h}\right]}_{x = 1} = - 22 + \setminus m a t h c a l \left\{O\right\} \left(h\right) = - 22$