# Is f(x)=1-x-e^(-3x)/x concave or convex at x=4?

Feb 5, 2018

Let's take some derivatives!

#### Explanation:

For $f \left(x\right) = 1 - x - {e}^{- 3 x} / x$, we have
$f ' \left(x\right) = - 1 - \frac{- 3 x {e}^{- 3 x} - {e}^{- 3 x}}{x} ^ 2$
This simplifies (sort of) to
$f ' \left(x\right) = - 1 + {e}^{- 3 x} \frac{3 x + 1}{x} ^ 2$
Therefore
$f ' ' \left(x\right) = {e}^{- 3 x} \frac{- 3 x - 2}{x} ^ 3 - 3 {e}^{- 3 x} \frac{3 x + 1}{x} ^ 2$
$= {e}^{- 3 x} \left(\frac{- 3 x - 2}{x} ^ 3 - 3 \frac{3 x + 1}{x} ^ 2\right)$
$= {e}^{- 3 x} \left(\frac{- 3 x - 2}{x} ^ 3 + \frac{- 9 x - 3}{x} ^ 2\right)$
$= {e}^{- 3 x} \left(\frac{- 3 x - 2}{x} ^ 3 + \frac{- 9 {x}^{2} - 3 x}{x} ^ 3\right)$
$= {e}^{- 3 x} \left(\frac{- 9 {x}^{2} - 6 x - 2}{x} ^ 3\right)$

Now let x = 4.

$f ' ' \left(4\right) = {e}^{- 12} \left(\frac{- 9 {\left(16\right)}^{2} - 6 \left(4\right) - 2}{4} ^ 3\right)$

Observe that the exponential is always positive. The numerator of the fraction is negative for all positive values of x. The denominator is positive for positive values of x.

Therefore $f ' ' \left(4\right) < 0$.