Question

What are the maximum and minimum values of the function `abs(2x-3` at closed intervals [1,2]?

what are the maximum and minimum values of the function `abs(2x-3` at closed intervals [1,2]

Answer 1

Please see below for a non-calculus answer.

Explanation:

Non-calculus answer

The graph the absolute value function is `V` shaped.

The graph of `f(x) = abs(2x-3)` is a translation and shrinking/stretching of the graph of `absx`. So it also has a `V` shape.

Therefore, the maximum value on a closed, bounded interval occurs at an endpoint of the interval.

For `f(x) = abs(2x-3)` on `[1,2]` the value of `f` is `1` at both endpoints, so the maximum is `1`.

Clearly `f(x) = abs(2x-3) >= 0` for all `x` and it is `0` only at `x = 3/2`.

Since `3/2` is in `[1,2]`, the minimum value of `f` on the interval `[1,2]` is `0` (this value occurs at `x=3/2`)

Answer 2

Please see below for a calculus answer.

Explanation:

Using the closed interval method for finding absolute extreme values:

`f(x) = abs(2x-3) = {(2x-3,"if", x >= 3/2),(-2x+3,"if", x < 3/2):}`

So `f'(x) = {(2,"if", x > 3/2),(-2,"if", x < 3/2):}`.

`f'(x)` is never `0` and `f'(3/2)` does not exist.

The only critical number for `f` is `3/2`.

Testing on the interval `{1,2}`

`{:(x," ",1," ",3/2," ",2),(f(x)," ",1," ",0," ",1):}`

The maximum is `1` and the minimum is `0`.