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

Feb 14, 2017

$f \downarrow , x \in \left(- \infty . - \frac{3}{2}\right] \mathmr{and} \left[10 , \frac{3}{2}\right]$.

$f \uparrow , x \in \left[- \frac{3}{2} , 10\right] \mathmr{and} \left[\frac{3}{2} , \infty\right)$.. See the Socratic depiction.

#### Explanation:

$y = f \left(x\right) \ge 1$. f is continuous, for $x \in \left(- \infty , \infty\right)$.

This is the combined equation to the truncated parabolas

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

${\left(x - \frac{3}{2}\right)}^{2} = \frac{1}{4} \left(y - 1\right)$, when $x \ge 0$ and

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

${\left(x + \frac{3}{2}\right)}^{2} = \frac{1}{4} \left(y - 1\right)$, when $x \le 0$.

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

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

one parabola changes to the other.
So,

$f \downarrow , x \in \left(- \infty . - \frac{3}{2}\right] \mathmr{and} \left[10 , \frac{3}{2}\right]$.

$f \uparrow , x \in \left[- \frac{3}{2} , 10\right] \mathmr{and} \left[\frac{3}{2} , \infty\right)$..

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