How do you determine the intervals for which the function is increasing or decreasing given $f (x) = {(2 | x | - 3)}^{2} + 1$ $f(x)=(2absx-3)^2+1$ ?

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Answer 1 · 2017-02-14T09:53:14

$f ⏐ ⏐ ⏐ ⏐ ↓, x \in (- \infty . - \frac{3}{2}] and [10, \frac{3}{2}]$ $f darr, x in (-oo. -3/2] and [10, 3/2]$ .

$f ↑ ⏐ ⏐ ⏐ ⏐, x \in [- \frac{3}{2}, 10] and [\frac{3}{2}, \infty)$ $f uarr, x in [-3/2, 10] and [3/2, oo)$ .#. See the Socratic depiction.

$y = f (x) \geq 1$ $y=f(x)>=1$ . f is continuous, for $x \in (- \infty, \infty)$ $x in (-oo, oo)$ .

This is the combined equation to the truncated parabolas

$y = {(2 x - 3)}^{2} + 1$ $y=(2x-3)^2+1$ , giving

${(x - \frac{3}{2})}^{2} = \frac{1}{4} (y - 1)$ $(x-3/2)^2=1/4(y-1)$ , when $x \geq 0$ $x >=0$ and

$y = {(2 x + 3)}^{2} + 1$ $y=(2x+3)^2+1$ , giving

${(x + \frac{3}{2})}^{2} = \frac{1}{4} (y - 1)$ $(x+3/2)^2=1/4(y-1)$ , when $x \leq 0$ $x <=0$ .

The vertices of these parabolas are at $(\pm \frac{3}{2}, 1)$ $(+-3/2, 1)$ .

The parabolas meet at (0, 10). This is the truncation point at which

one parabola changes to the other.
So,

$f ⏐ ⏐ ⏐ ⏐ ↓, x \in (- \infty . - \frac{3}{2}] and [10, \frac{3}{2}]$ $f darr, x in (-oo. -3/2] and [10, 3/2]$ .

$f ↑ ⏐ ⏐ ⏐ ⏐, x \in [- \frac{3}{2}, 10] and [\frac{3}{2}, \infty)$ $f uarr, x in [ -3/2, 10] and [3/2, oo)$ .#.

graph{((x-3/2)^2-1/4(y-1))((x+3/2)^2-1/4(y-1))=0 [-10, 10, 0, 10]}

How do you determine the intervals for which the function is increasing or decreasing given f(x)=(2absx-3)^2+1f(x)=(2|x|−3)2+1f(x)=(2absx-3)^2+1?