Question

How do you find the range of `y = 1 + (x/(x^2+1))`?

Answer 1

The function is defined for all real `x`, so let's find the extrema by differentiating and the setting the derivative equal to 0.

Differentiate:

`dy/dx = (d(1))/dx+(d(x/(x^2+1)))/dx`

The derivative of a constant is 0:

`dy/dx = 0+(d(x/(x^2+1)))/dx`

`dy/dx = (d(x/(x^2+1)))/dx`

Using the quotient rule:

`dy/dx = ((d(x))/dx(x^2+1)-x(d(x^2+1))/dx)/(x^2+1)^2`

`dy/dx = (x^2+1-x(2x))/(x^2+1)^2`

`dy/dx = (x^2+1-2x^2)/(x^2+1)^2`

`dy/dx = (1-x^2)/(x^2+1)^2`

Now that we have the first derivative, we shall set it equal to 0 and solve for the value(s) of x:

`0 = (1-x^2)/(x^2+1)^2`

Since `(x^2+1)^2≠0` for all `x`,
`0 = 1-x^2`

`x^2=1`

`x = +-1`

We can find the range by evaluating the function at `x = -1` and `x = 1`:

`y = 1 + (-1)/((-1)^2+1)` and `y = 1 + (1)/(1^2+1)`

`y = 1/2` and `y = 3/2`

The range is:

`1/2<= y <= 3/2`

We can verify this by graphing the equation:

graph{y = 1 + (x/(x^2+1)) [-10, 10, -5, 5]}