# What are the local extrema, if any, of f(x) =2x^3 -3x^2+7x-2 ?

May 25, 2017

Are no local extremas in ${\mathbb{R}}^{n}$ for $f \left(x\right)$

#### Explanation:

We'll first need to take the derivative of $f \left(x\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] - 3 \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] + 7 \frac{d}{\mathrm{dx}} \left[x\right] - 0$
$= 6 {x}^{2} - 6 x + 7$

So, $f ' \left(x\right) = 6 {x}^{2} - 6 x + 7$

To solve for the local extremas, we must set the derivative to $0$

$6 {x}^{2} - 6 x + 7 = 0$
$x = \frac{6 \pm \sqrt{{6}^{2} - 168}}{12}$

Now, we've hit a problem. It's that $x \in \mathbb{C}$ so the local extremas are complex. This is what happens when we start off in cubic expressions, it's that complex zeros can happen in the first derivative test. In this case, there are no local extremas in ${\mathbb{R}}^{n}$ for $f \left(x\right)$.