# Question #115fc

Jul 14, 2017

Maximum at $\left(- 8 , 75\right)$

#### Explanation:

We have:

$g \left(x\right) = - {x}^{2} - 16 x + 1$

Whose graph is as follows:
graph{-x^2-16x+1 [-25, 10, -20, 80]}

General Observations

$g \left(x\right)$ is a quadratic so we have a parabola with a single turning point. The ${x}^{2}$ coefficient is negative so the parabola is inverted ($\cap$ shaped) and so we have a single maximum .

We can demonstrate this is the case using two methods:

Method 1 - Completing the Square

$g \left(x\right) = - \left({x}^{2} + 16 x - 11\right)$
$\text{ } = - \left({\left(x + \frac{16}{2}\right)}^{2} - {\left(\frac{16}{2}\right)}^{2} - 11\right)$
$\text{ } = - \left({\left(x + 8\right)}^{2} - {8}^{2} - 11\right)$
$\text{ } = - \left({\left(x + 8\right)}^{2} - 64 - 11\right)$
$\text{ } = - \left({\left(x + 8\right)}^{2} - 75\right)$
$\text{ } = - {\left(x + 8\right)}^{2} + 75$

Clearly ${\left(x + 8\right)}^{2} \ge 0 \forall x \in \mathbb{R}$, and $x + 8 = 0 \implies x = - 8$ and so $g \left(x\right)$ has a maximum of $75$ when $x = - 8$

Method 2 - Calculus

Differentiating (twice) we get:

$\setminus g ' \left(x\right) = - 2 x - 16$
$g ' ' \left(x\right) = - 2$

At a critical point (min or max) the first derivative vanishes, thus:

$g ' \left(x\right) = - 2 x - 16 = 0 \implies - 2 x - 16 = 0$
$\therefore 2 x = - 16 \implies x = - 8$

So there is one critical point when $x = - 8$

The nature of the critical point is determined by the sign of the second derivative:

$x = - 8 \implies g ' ' \left(- 8\right) = - 2 < 0$

And as $g ' ' \left(x\right) < 0$ the critical point is a maximum .