Question

What are the critical values, if any, of `f(x) = (x - 1) / (x + 3) -sqrt(x^2-3)`?

Answer 1

There are critical values at:

`x = +-sqrt(3)`
`x = -4.935`

Explanation:

Start by finding the derivative using the quotient and chain rules.

`f'(x) = (1(x + 3) - 1(x - 1))/(x + 3)^2 - (2x)/(2sqrt(x^2 - 3)`

`f'(x) = (x + 3 - x + 1)/(x + 3)^2 - x/sqrt(x^2 - 3)`

`f'(x) = 4/(x + 3)^2 - x/sqrt(x^2 - 3)`

Critical points will occur whenever the derivative equals `0` (a horizontal tangent or is undefined (vertical tangent).

The derivative will be undefined whenever the denominator equals `0`.

`(x + 3)^2 = 0`

`x = -3`

AND

`sqrt(x^2 - 3) = 0`

`x = +- sqrt(3)`

However, we disregard the first critical value, `x= -3`, because it also renders the function undefined.

Now let's see where the derivative equals `0`.

`0 = 4/(x +3)^2 - x/sqrt(x^2 - 3)`

`x/sqrt(x^2 - 3) = 4/(x + 3)^2`

`x^2/(x^2 - 3) = 16/(x + 3)^4`

`x^2(x + 3)^4 = 16(x^2 - 3)`

This equation is very difficult, if not impossible to solve algebraically. Use either Newton's Method, a CAS or a graphing calculator to solve.

I'll use the latter. On your equations tab, enter

`{(y_1 = x^2(x + 3)^4), (y_2 = 16(x^2 - 3)):}`

Find the intersect-finding function. These will give you your solutions. This will be `x = -1.839` and `x = -4.935`. However, `x = -1.839` is extraneous, so it is not a critical value.

So, you're done here!

Hopefully this helps!