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

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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]}