Calculus

[wicked juniper]

Is this function continuous and differentiable?
$f(x) = \abs{x^2-2x+1}$

[wicked juniper]

$f(x) = \abs{x^2-2x+1}$

The bost isn't working

f(x) = |x²-2x+1|

[gleaming olive]

this function never really crosses below the x-axis, so that means it doesn't matter if there is an absolute value there or not

in other words, |x²-2x+1|=x²-2x+1

[wicked juniper]

No because it's going to split to two parts

[gleaming olive]

What do you mean by split in 2 parts?

Here let me put it this way, if we find the roots of x^2-2x+1, we notice that it only has 1 root. This means that it doesn't go below the x-axis

It touches the x axis

This means that any value we plug into x, y>=0

So if we take the absolute value of the equation, we will always be taking the absolute value of a positive number

Which is just a positive

[wicked juniper]

This is what i mean

Is this function continuous when x=3

So what we normally do

[gleaming olive]

Those are both the same equation, just one has a negative distributed to it

[wicked juniper]

Find f(3)

Then lim x→3 from the right

Then lim x→3 from the left

[wicked juniper]

I know

If they're equal then the function is continuous

Anyways i did it and thank you

.solved