# What are the local extrema, if any, of f(x) =x^2 + 9x +1 ?

Mar 8, 2016

Parabolae have exactly one extrema, the vertex.

It is $\left(- 4 \frac{1}{2} , - 19 \frac{1}{4}\right)$.

Since $\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} = 2$ everywhere the function is concave up everywhere and this point must be a minimum.

#### Explanation:

You have two roots to finding the vertex of the parabola: one, use calculus to find were the derivative is zero; two, avoid calculus at all costs and just complete the square. We're going to use calculus for the practice.

$f \left(x\right) = {x}^{2} + 9 x + 1$, we need to take the derivative of this.

$\frac{d f \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{2} + 9 x + 1\right)$

By the linearity of the derivative we have

$\frac{d f \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(9 x\right) + \frac{d}{\mathrm{dx}} \left(1\right)$.

Using the power rule, $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$ we have

$\frac{d f \left(x\right)}{\mathrm{dx}} = 2 \cdot {x}^{1} + 9 \cdot 1 \cdot {x}^{0} + 0 = 2 x + 9$.

We set this equal to zero to find the critical points, the local and global minima and maxima and sometimes points of inflection have derivatives of zero.

$0 = 2 x + 9$ $\implies$ $x = - \frac{9}{2}$,
so we have one critical point at $x = - \frac{9}{2}$ or $- 4 \frac{1}{2}$.

To find the y coordinate of the critical point we sub in $x = - \frac{9}{2}$ back into the function,

$f \left(- \frac{9}{2}\right) = {\left(- \frac{9}{2}\right)}^{2} + 9 \left(- \frac{9}{2}\right) + 1 = \frac{81}{4} - \frac{81}{2} + 1$
$= \frac{81}{4} - \frac{162}{4} + \frac{4}{4} = - \frac{77}{4} = - 19 \frac{1}{4}$.

The critical point/vertex is $\left(- 4 \frac{1}{2} , - 19 \frac{1}{4}\right)$.

We know that because $a > 0$, this is a maximum.
To formally find if it's a maxima or minima we need to do the second derivative test.

$\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(2 x + 9\right) = \frac{d}{\mathrm{dx}} \left(2 x\right) + \frac{d}{\mathrm{dx}} \left(9\right) = 2 + 0 = 2$

The second derivative is 2 at all values of x. This means it is greater then zero everywhere, and the function is concave up everywhere (it's a parabola with $a > 0$ after all), so the extrema must be a minimum, the vertex.