Finding the Range Algebraically

[Math Tutor]

Find the range of each function using algebra only.

(a) [tex]f(x)=\dfrac{3x+2}{x-4}[/tex]

(b) [tex]g(x)=\dfrac{x}{x^{2}+1}[/tex]

(c) [tex]h(x)=\dfrac{x^{2}-x+1}{x^{2}+x+1}[/tex]

(d) [tex]k(x)=3-\sqrt{2x-5}[/tex]

[nycmath]

Find the range of each function using algebra only.

(a) [tex]f(x)=\dfrac{3x+2}{x-4}[/tex]

(b) [tex]g(x)=\dfrac{x}{x^{2}+1}[/tex]

(c) [tex]h(x)=\dfrac{x^{2}-x+1}{x^{2}+x+1}[/tex]

(d) [tex]k(x)=3-\sqrt{2x-5}[/tex]

I will work on these tomorrow. Thanks.

[nycmath]

(a) f(x) = (3x + 2)/(x - 4)

Let f(x) = y

y = (3x + 2)/(x - 4)

Switch x and y.

y(x - 4) = 3x + 2

xy - 4y = 3x + 2

xy - 3x = 4y + 2

x(y - 3)/(y - 3) = (3y + 2)/(y - 3)

x = (3y + 2)/(y - 3)

Set y - 3 = 0.

y = 3

Range = {y | y cannot be 3}

Yes?

So, the range of the original function is the domain of the inverse of the original function.

Yes?

Note: I will do the rest later today.

[Eigenvalue]

(a) f(x) = (3x + 2)/(x - 4)

Let f(x) = y

y = (3x + 2)/(x - 4)

Switch x and y.

y(x - 4) = 3x + 2

xy - 4y = 3x + 2

xy - 3x = 4y + 2

x(y - 3)/(y - 3) = (3y + 2)/(y - 3)

x = (3y + 2)/(y - 3)

Set y - 3 = 0.

y = 3

Range = {y | y cannot be 3}

Yes?

So, the range of the original function is the domain of the inverse of the original function.

Yes?

Note: I will do the rest later today.

Yes,.because f(g(x)=g(f(x)=x, if f(x) and g(x) are inverse functions

[nycmath]

(a) f(x) = (3x + 2)/(x - 4)

Let f(x) = y

y = (3x + 2)/(x - 4)

Switch x and y.

y(x - 4) = 3x + 2

xy - 4y = 3x + 2

xy - 3x = 4y + 2

x(y - 3)/(y - 3) = (3y + 2)/(y - 3)

x = (3y + 2)/(y - 3)

Set y - 3 = 0.

y = 3

Range = {y | y cannot be 3}

Yes?

So, the range of the original function is the domain of the inverse of the original function.

Yes?

Note: I will do the rest later today.

Yes,.because f(g(x)=g(f(x)=x, if f(x) and g(x) are inverse functions

I will do the rest now.

[nycmath]

See attachment.

[nycmath]

See attachment.

[nycmath]

See attachment.

[Math Tutor]

All three final answers are correct: (b) -1/2 <= y <= 1/2, (c) 1/3 <= y <= 3, (d) (-inf, 3].

Two places where the reasoning should be tightened, though.

1) The discriminant test is only valid when the coefficient of x^2 is not zero, so that one value of y has to be checked separately.
- In (b) you have a = y, so y = 0 is not covered by D >= 0. Check it directly: y = 0 gives -x = 0, so x = 0, and indeed g(0) = 0. So 0 is in the range (and it sits inside your interval anyway).
- In (c) you have a = y - 1, so check y = 1 by hand: the equation becomes 2x = 0, x = 0, and h(0) = 1/1 = 1. Again it belongs to the range, so your interval is unchanged.

2) In (b), the step sqrt(y^2) <= sqrt(1/4) => y <= +/- 1/2 is not legitimate as written; "y <= +/- 1/2" has no meaning. Since sqrt(y^2) = |y|, the line should read |y| <= 1/2, which is exactly -1/2 <= y <= 1/2. Same answer, but now the reasoning is airtight.

(d) is fine as it stands. You can even say it in one line: as x runs over [5/2, inf), sqrt(2x-5) runs over all of [0, inf), hence 3 - sqrt(2x-5) runs over all of (-inf, 3]. The maximum value 3 is attained at x = 5/2.

[nycmath]

All three final answers are correct: (b) -1/2 <= y <= 1/2, (c) 1/3 <= y <= 3, (d) (-inf, 3].

Two places where the reasoning should be tightened, though.

1) The discriminant test is only valid when the coefficient of x^2 is not zero, so that one value of y has to be checked separately.
- In (b) you have a = y, so y = 0 is not covered by D >= 0. Check it directly: y = 0 gives -x = 0, so x = 0, and indeed g(0) = 0. So 0 is in the range (and it sits inside your interval anyway).
- In (c) you have a = y - 1, so check y = 1 by hand: the equation becomes 2x = 0, x = 0, and h(0) = 1/1 = 1. Again it belongs to the range, so your interval is unchanged.

2) In (b), the step sqrt(y^2) <= sqrt(1/4) => y <= +/- 1/2 is not legitimate as written; "y <= +/- 1/2" has no meaning. Since sqrt(y^2) = |y|, the line should read |y| <= 1/2, which is exactly -1/2 <= y <= 1/2. Same answer, but now the reasoning is airtight.

(d) is fine as it stands. You can even say it in one line: as x runs over [5/2, inf), sqrt(2x-5) runs over all of [0, inf), hence 3 - sqrt(2x-5) runs over all of (-inf, 3]. The maximum value 3 is attained at x = 5/2.

I like looking for the range algebraically far more than using the graph of functions. I will need, however, to become proficient with both techniques.