# How do you find the critical points and local max and min for y=(x-1)^4?

Dec 6, 2017

Min: $x = 1$

Explained mathematically but easy to see graphically.

#### Explanation:

$y = {\left(x - 1\right)}^{4}$
Power rule + chain rule
$y ' = 4 {\left(x - 1\right)}^{3} \cdot 1$
$y ' = 4 {\left(x - 1\right)}^{3}$

Critical points are when $f ' \left(x\right) = 0$ so...
we set $y ' = 0$

$0 = 4 {\left(x - 1\right)}^{3}$
$\frac{0}{\textcolor{red}{4}} = \frac{4 {\left(x - 1\right)}^{3}}{\textcolor{red}{4}}$

$0 = {\left(x - 1\right)}^{3}$
0^(color(red)(1/3)=((x-1)^3)^color(red)(1/3)

$0 \textcolor{red}{+} \textcolor{red}{1} = x - 1 \textcolor{red}{+} \textcolor{red}{1}$
$1 = x$

We now know that $x = 1$ is a critical point.

Now we look at the original function $y = {\left(x - 1\right)}^{4}$
We can take a look at $x = 1$ and compare that to $x$ slightly smaller than 1 and $x$ slightly larger than 1

$y = {\left(1 - 1\right)}^{4}$
$y = {\left(0\right)}^{4}$
$y = 0$

$y = {\left(0.99 - 1\right)}^{4}$
$y = {\left(- 0.01\right)}^{4}$
$y = 0.00000001$

$y = {\left(1.01 - 1\right)}^{4}$
$y = {\left(.01\right)}^{4}$
$y = 0.00000001$

The points immediately to the left and right of $x = 1$ are both larger than $x = 1$ therefore $x = 1$ must be a minimum