Respuesta :
as you'd know, to get the inverse, we'd do a quick switcharoo on the variables first off, and then solve for "y".
hmmmm the restrictions on the original, I the domain are none, on the range well, if we put the quadratic in "vertex form"
[tex]\bf h(x)=x^2+6x+9\implies h(x)=(x+3)(x+3)\implies h(x)=(x+3)^2 \\\\\\ \stackrel{vertex~form}{h(x)}=[x-\stackrel{h}{(-3)}]^2\stackrel{k}{+0}\impliedby vertex~(h,k)[/tex]
so, notice, the vertex of the parabola is at (-3, 0), what does that mean for the range?
well, it means, the graph comes from above and goes down down down, reaches -3,0, makes a U-turn, then goes back up up up.
so, the lowest value over the y-axis is 0, it never goes below that, so the constraints for the range is y ⩾ 0.
now, let's do the inverse,
[tex]\bf h(x)=x^2+6x+9\implies h(x)=(x+3)(x+3)\implies \stackrel{h(x)}{y}=(x+3)^2 \\\\\\ \boxed{x}=\left(\boxed{y}+3 \right)^2\impliedby inverse\implies \pm\sqrt{x}=y+3 \\\\\\ \pm\sqrt{x}-3=\stackrel{f^{-1}(x)}{y}[/tex]
now, what are the constraints for the inverse? well, recall that, for an inverse, its domain is the same as the range of the original, and its range is the domain of the original.
so, we know the domain for the original has no constraints, thus the range of the inverse has no constraints, since its the same set anyway.
now, the original has y ⩾ 0 as constraint on its range, thus the domain of the inverse will run into the same constraint of y ⩾ 0.
hmmmm the restrictions on the original, I the domain are none, on the range well, if we put the quadratic in "vertex form"
[tex]\bf h(x)=x^2+6x+9\implies h(x)=(x+3)(x+3)\implies h(x)=(x+3)^2 \\\\\\ \stackrel{vertex~form}{h(x)}=[x-\stackrel{h}{(-3)}]^2\stackrel{k}{+0}\impliedby vertex~(h,k)[/tex]
so, notice, the vertex of the parabola is at (-3, 0), what does that mean for the range?
well, it means, the graph comes from above and goes down down down, reaches -3,0, makes a U-turn, then goes back up up up.
so, the lowest value over the y-axis is 0, it never goes below that, so the constraints for the range is y ⩾ 0.
now, let's do the inverse,
[tex]\bf h(x)=x^2+6x+9\implies h(x)=(x+3)(x+3)\implies \stackrel{h(x)}{y}=(x+3)^2 \\\\\\ \boxed{x}=\left(\boxed{y}+3 \right)^2\impliedby inverse\implies \pm\sqrt{x}=y+3 \\\\\\ \pm\sqrt{x}-3=\stackrel{f^{-1}(x)}{y}[/tex]
now, what are the constraints for the inverse? well, recall that, for an inverse, its domain is the same as the range of the original, and its range is the domain of the original.
so, we know the domain for the original has no constraints, thus the range of the inverse has no constraints, since its the same set anyway.
now, the original has y ⩾ 0 as constraint on its range, thus the domain of the inverse will run into the same constraint of y ⩾ 0.
