How do you find the critical points and local max and min for y=(x-1)^4y=(x1)4?

1 Answer
Dec 6, 2017

Min: x=1x=1

Explained mathematically but easy to see graphically.

Explanation:

y = (x-1)^4y=(x1)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