How do use the first derivative test to determine the local extrema x^2-x-1?

Sep 12, 2015

$\left(\frac{1}{2} , - \frac{5}{4}\right)$.

Explanation:

The "peaks" or local extrema of a function $f \left(x\right)$ occur at the values where $\frac{d}{\mathrm{dx}} f \left(x\right) = 0$.

One way to remember this is:
Since a function is increasing when $\frac{d}{\mathrm{dx}} f \left(x\right) > 0$, and a function is decreasing when $\frac{d}{\mathrm{dx}} f \left(x\right) < 0$, that means when $\frac{d}{\mathrm{dx}} f \left(x\right) = 0$, the graph of the function is "turning" from increasing to decreasing or from decreasing to increasing. The turn forms a "peak" which is a local extrema.

So, to get the values for the local extrema for $f \left(x\right) = {x}^{2} - x - 1$, we need to evaluate $f \left(x\right)$ at the values of $x$ where $\frac{d}{\mathrm{dx}} f \left(x\right) = 0$.

First, we get the derivative using the power rule:
$\textcolor{g r e e n}{\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}}$

$f \left(x\right) = {x}^{2} - x - 1$

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

Then, we solve for the values where $\frac{d}{\mathrm{dx}} f \left(x\right) = 0$:

$0 = 2 x - 1$
$1 = 2 x$
$\frac{1}{2} = x$

So, there is a local extrema when $x = \frac{1}{2}$.

To find the value of the local extrema, we evaluate $f \left(\frac{1}{2}\right)$.

$f \left(x\right) = {x}^{2} - x - 1$
$f \left(\frac{1}{2}\right) = {\left(\frac{1}{2}\right)}^{2} - \frac{1}{2} - 1$
$f \left(\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 1$
$f \left(\frac{1}{2}\right) = - \frac{5}{4}$

There exists a local extrema at the point $\left(\frac{1}{2} , - \frac{5}{4}\right)$.