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

Question

1610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-14T09:53:14

$f darr, x in (-oo. -3/2] and [10, 3/2]$ .

$f uarr, x in [-3/2, 10] and [3/2, oo)$ .#. See the Socratic depiction.

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

This is the combined equation to the truncated parabolas

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

$(x-3/2)^2=1/4(y-1)$ , when $x >=0$ and

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

$(x+3/2)^2=1/4(y-1)$ , when $x <=0$ .

The vertices of these parabolas are at $(+-3/2, 1)$ .

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

one parabola changes to the other.
So,

$f darr, x in (-oo. -3/2] and [10, 3/2]$ .

$f uarr, x in [ -3/2, 10] and [3/2, oo)$ .#.

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)=(2absx-3)^2+1?