Question

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

Answer 1

Min: `x=1`

Explained mathematically but easy to see graphically.

Explanation:

`y = (x-1)^4`
Power rule + chain rule
`y' = 4(x-1)^3 * 1`
`y' = 4(x-1)^3`

Critical points are when `f'(x) = 0` so...
we set `y' = 0`

`0 = 4(x-1)^3`
`0/color (red) 4 = (4(x-1)^3)/color(red)4`

`0 = (x-1)^3`
`0^(color(red)(1/3)=((x-1)^3)^color(red)(1/3)`

`0color(red)+color(red)1 = x - 1 color(red) +color(red)1`
`1 = x`

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

Now we look at the original function `y = (x-1)^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 = (1-1)^4`
`y = (0)^4`
`y = 0`

`y = (0.99-1)^4`
`y = (-0.01)^4`
`y = 0.00000001`

`y = (1.01-1)^4`
`y = (.01)^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