# How do you find the coordinates of the local extrema of the function?

Sep 7, 2014

The best way to do this is to find the derivative of the function.

Let's say our function is $f \left(x\right) = \frac{3}{7} {x}^{2} - 7 x + 13$. Not a particularly pretty or elegant function, but we can still find its minimum.

First, we take the derivative of the function. Remember:
$\text{if "f(x)=x^a," then } f ' \left(x\right) = a {x}^{a - 1}$
So: $f ' \left(x\right) = \frac{6}{7} x - 7$

Now we need to find the point where $f ' \left(x\right) = 0$, i.e. the slope of the original function is 0.

So $\frac{6}{7} x - 7 = 0$
$\implies \frac{6}{7} x = 7$
$\implies 6 x = 49$
$\implies x = \frac{49}{6}$

The last step is to plug this x value into the original equation.

$f \left(\frac{49}{6}\right) = \frac{3}{7} {\left(\frac{49}{6}\right)}^{2} - 7 \left(\frac{49}{6}\right) + 13$
$= \frac{3}{7} \cdot \frac{2401}{36} - \frac{2401}{6} + 13$
$= \frac{343}{12} - \frac{343}{6} + 13$
$= \frac{343}{12} - \frac{686}{12} + \frac{156}{12}$
$= - \frac{187}{12}$

So the local minimum is at $\left(\frac{49}{6} , - \frac{187}{12}\right)$.

Note that I deliberately didn't pick a "nice and easy" function - this is to show that finding the derivative, and making the derivative equal zero, works for everything.