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

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

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

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Answer 1 · 2018-03-28T16:22:51

Please see below for a non-calculus answer.

Explanation:

Non-calculus answer

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

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

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

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

Clearly $f (x) = | 2 x - 3 | \geq 0$ $f(x) = abs(2x-3) >= 0$ for all $x$ $x$ and it is $0$ $0$ only at $x = \frac{3}{2}$ $x = 3/2$ .

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

Answer 2 · 2018-03-28T16:29:56

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 $0$ .

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