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

1 Answer
Mar 1, 2017

Answer:

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!