Find Domain & Range Algebraically

[nycmath]

See attachment. Show your steps by explaining what you are doing along the way to the right answer.

[Guest]

Hi its Bob,

The domain is the easy half — the only thing that can break is a zero denominator, so set [tex](x-1)(x+1)=0[/tex] and throw those two values out. The domain is every real number except [tex]x=1[/tex] and [tex]x=-1[/tex].

For the range, turn the problem around and solve for [tex]x[/tex] in terms of [tex]y[/tex].

[tex]y(x^2-1)=3x-2[/tex], so [tex]yx^2-3x+(2-y)=0[/tex].

If [tex]y=0[/tex], that drops to [tex]-3x+2=0[/tex], giving [tex]x=\frac{2}{3}[/tex], so [tex]y=0[/tex] is hit.

If [tex]y\neq 0[/tex] it is a quadratic in [tex]x[/tex], and it has a real solution exactly when the discriminant is not negative:

[tex]D=9-4y(2-y)=4y^2-8y+9[/tex].

That one has discriminant [tex]64-144=-80<0[/tex] and opens upward, so [tex]D>0[/tex] for every [tex]y[/tex]. And the [tex]x[/tex] you get is never [tex]\pm 1[/tex] — putting [tex]x=1[/tex] in the quadratic leaves [tex]-1[/tex] and [tex]x=-1[/tex] leaves [tex]5[/tex], never zero.

So every [tex]y[/tex] is reachable: the range is all real numbers.

[nycmath]

Hi its Bob,

The domain is the easy half — the only thing that can break is a zero denominator, so set [tex](x-1)(x+1)=0[/tex] and throw those two values out. The domain is every real number except [tex]x=1[/tex] and [tex]x=-1[/tex].

For the range, turn the problem around and solve for [tex]x[/tex] in terms of [tex]y[/tex].

[tex]y(x^2-1)=3x-2[/tex], so [tex]yx^2-3x+(2-y)=0[/tex].

If [tex]y=0[/tex], that drops to [tex]-3x+2=0[/tex], giving [tex]x=\frac{2}{3}[/tex], so [tex]y=0[/tex] is hit.

If [tex]y\neq 0[/tex] it is a quadratic in [tex]x[/tex], and it has a real solution exactly when the discriminant is not negative:

[tex]D=9-4y(2-y)=4y^2-8y+9[/tex].

That one has discriminant [tex]64-144=-80<0[/tex] and opens upward, so [tex]D>0[/tex] for every [tex]y[/tex]. And the [tex]x[/tex] you get is never [tex]\pm 1[/tex] — putting [tex]x=1[/tex] in the quadratic leaves [tex]-1[/tex] and [tex]x=-1[/tex] leaves [tex]5[/tex], never zero.

So every [tex]y[/tex] is reachable: the range is all real numbers.

I like finding the range algebraically more so than using a graph.