A Function and Its Inverse

[Math Tutor]

Let [tex]f(x)=\dfrac{2x-3}{x+1}[/tex].

(a) Give the domain and the range of [tex]f[/tex].
(b) Show that [tex]f[/tex] is one-to-one and find [tex]f^{-1}(x)[/tex].
(c) Compare the domain and range of [tex]f^{-1}[/tex] with those of [tex]f[/tex]. What do you notice?
(d) Solve [tex]f(x)=f^{-1}(x)[/tex].

Part (d) is the interesting one. Most students start by setting [tex]f(x)=x[/tex], on the idea that the two graphs can only meet on the line [tex]y=x[/tex]. Try it the direct way instead and see what the algebra tells you about where the two curves cross.

[nycmath]

Let [tex]f(x)=\dfrac{2x-3}{x+1}[/tex].

(a) Give the domain and the range of [tex]f[/tex].
(b) Show that [tex]f[/tex] is one-to-one and find [tex]f^{-1}(x)[/tex].
(c) Compare the domain and range of [tex]f^{-1}[/tex] with those of [tex]f[/tex]. What do you notice?
(d) Solve [tex]f(x)=f^{-1}(x)[/tex].

Part (d) is the interesting one. Most students start by setting [tex]f(x)=x[/tex], on the idea that the two graphs can only meet on the line [tex]y=x[/tex]. Try it the direct way instead and see what the algebra tells you about where the two curves cross.

See attachment.

[nycmath]

Let [tex]f(x)=\dfrac{2x-3}{x+1}[/tex].

(a) Give the domain and the range of [tex]f[/tex].
(b) Show that [tex]f[/tex] is one-to-one and find [tex]f^{-1}(x)[/tex].
(c) Compare the domain and range of [tex]f^{-1}[/tex] with those of [tex]f[/tex]. What do you notice?
(d) Solve [tex]f(x)=f^{-1}(x)[/tex].

Part (d) is the interesting one. Most students start by setting [tex]f(x)=x[/tex], on the idea that the two graphs can only meet on the line [tex]y=x[/tex]. Try it the direct way instead and see what the algebra tells you about where the two curves cross.

See attachment for parts (c) and (d).

[Math Tutor]

Good work — (b) is completely correct, and the algebra in (d) is right as well. Two things to fix.

(a) You wrote "Domain: set [tex]x+1=0[/tex]". It is the other way round: you require [tex]x+1\neq 0[/tex]; solving [tex]x+1=0[/tex] only locates the value you must throw out. The answer [tex]D_f=\mathbb{R}\setminus\{-1\}[/tex] is correct.

For the range you do not need a graph. Divide:
[tex]f(x)=\frac{2x-3}{x+1}=2-\frac{5}{x+1}[/tex]
and since [tex]\frac{5}{x+1}[/tex] is never [tex]0[/tex], the value [tex]2[/tex] is never taken, so [tex]R_f=\mathbb{R}\setminus\{2\}[/tex].
Equivalently, solve [tex]y=\frac{2x-3}{x+1}[/tex] for [tex]x[/tex]: [tex]x(y-2)=-y-3[/tex], which is solvable exactly when [tex]y\neq 2[/tex]. Same conclusion — and this is also where your [tex]f^{-1}[/tex] comes from.

(b) Correct: cross-multiplying [tex]\frac{2a-3}{a+1}=\frac{2b-3}{b+1}[/tex] gives [tex]5a=5b[/tex], so [tex]a=b[/tex] and [tex]f[/tex] is one-to-one, and
[tex]f^{-1}(x)=\frac{-x-3}{x-2}=\frac{x+3}{2-x}[/tex]
Yes — that is right.

(c) With the range in hand this is immediate:
[tex]D_{f^{-1}}=\mathbb{R}\setminus\{2\}=R_f[/tex] and [tex]R_{f^{-1}}=\mathbb{R}\setminus\{-1\}=D_f[/tex]
The domain and the range swap. That always happens for an inverse, and here you can see it explicitly.

(d) Your trinomial is right: [tex](2x-3)(2-x)=(x+3)(x+1)[/tex] gives [tex]3x^2-3x+9=0[/tex], i.e. [tex]x^2-x+3=0[/tex], and the quadratic formula does give [tex]\frac{1\pm i\sqrt{11}}{2}[/tex].

But check the discriminant first: [tex]D=1-12=-11<0[/tex]. That one line already settles it. Here [tex]f[/tex] is a real function with domain [tex]\mathbb{R}\setminus\{-1\}[/tex], and [tex]\frac{1\pm i\sqrt{11}}{2}[/tex] does not belong to it, so the equation [tex]f(x)=f^{-1}(x)[/tex] has no solution: the graphs of [tex]f[/tex] and [tex]f^{-1}[/tex] never meet. Reporting the complex roots answers a different question from the one that was asked.

One last remark: [tex]f(x)=x[/tex] leads to the very same equation [tex]x^2-x+3=0[/tex], so the popular shortcut agrees with the direct method here. Do not trust it in general though — a decreasing function can meet its inverse at points off the line [tex]y=x[/tex].

[nycmath]

Good work — (b) is completely correct, and the algebra in (d) is right as well. Two things to fix.

(a) You wrote "Domain: set [tex]x+1=0[/tex]". It is the other way round: you require [tex]x+1\neq 0[/tex]; solving [tex]x+1=0[/tex] only locates the value you must throw out. The answer [tex]D_f=\mathbb{R}\setminus\{-1\}[/tex] is correct.

For the range you do not need a graph. Divide:
[tex]f(x)=\frac{2x-3}{x+1}=2-\frac{5}{x+1}[/tex]
and since [tex]\frac{5}{x+1}[/tex] is never [tex]0[/tex], the value [tex]2[/tex] is never taken, so [tex]R_f=\mathbb{R}\setminus\{2\}[/tex].
Equivalently, solve [tex]y=\frac{2x-3}{x+1}[/tex] for [tex]x[/tex]: [tex]x(y-2)=-y-3[/tex], which is solvable exactly when [tex]y\neq 2[/tex]. Same conclusion — and this is also where your [tex]f^{-1}[/tex] comes from.

(b) Correct: cross-multiplying [tex]\frac{2a-3}{a+1}=\frac{2b-3}{b+1}[/tex] gives [tex]5a=5b[/tex], so [tex]a=b[/tex] and [tex]f[/tex] is one-to-one, and
[tex]f^{-1}(x)=\frac{-x-3}{x-2}=\frac{x+3}{2-x}[/tex]
Yes — that is right.

(c) With the range in hand this is immediate:
[tex]D_{f^{-1}}=\mathbb{R}\setminus\{2\}=R_f[/tex] and [tex]R_{f^{-1}}=\mathbb{R}\setminus\{-1\}=D_f[/tex]
The domain and the range swap. That always happens for an inverse, and here you can see it explicitly.

(d) Your trinomial is right: [tex](2x-3)(2-x)=(x+3)(x+1)[/tex] gives [tex]3x^2-3x+9=0[/tex], i.e. [tex]x^2-x+3=0[/tex], and the quadratic formula does give [tex]\frac{1\pm i\sqrt{11}}{2}[/tex].

But check the discriminant first: [tex]D=1-12=-11<0[/tex]. That one line already settles it. Here [tex]f[/tex] is a real function with domain [tex]\mathbb{R}\setminus\{-1\}[/tex], and [tex]\frac{1\pm i\sqrt{11}}{2}[/tex] does not belong to it, so the equation [tex]f(x)=f^{-1}(x)[/tex] has no solution: the graphs of [tex]f[/tex] and [tex]f^{-1}[/tex] never meet. Reporting the complex roots answers a different question from the one that was asked.

One last remark: [tex]f(x)=x[/tex] leads to the very same equation [tex]x^2-x+3=0[/tex], so the popular shortcut agrees with the direct method here. Do not trust it in general though — a decreasing function can meet its inverse at points off the line [tex]y=x[/tex].

Thank you for correcting my work. I need more work on finding the range algebraically.