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

Question

Ezra Adams · Accepted Answer

The critical points for #f(x)# are

#{-1/2, (-1+sqrt(13))/2, (-1-sqrt(13))/2}# The critical points of a function #f(x)# are the points #c# at which #f'(c) = 0# or #f'(c)# is undefined. Thus first we find the derivative of the function, and then set it equal to zero. To find the derivative we apply the quotient rule, the chain rule, and the power rule: #f'(x) = d/dx1/sqrt(x+x^2-3)# #= -(1/2(x+x^2-3)^(-1/2)(2x+1))/(x+x^2-3)# #=-(2x+1)/(2(x+x^2-3)^(3/2))# Now, to find the critical points, we want to find all points where #f'# is #0# or undefined. Because #f'(x) = 0 <=> 2x+1 = 0# we can solve for #x# to find a critical point at #-1/2#. As #f'(x)# is undefined if and only if #2(x+x^2 - 3)^(3/2) = 0 <=> x^2 + x - 3 = 0# we can use the quadratic formula to find critical points at #(-1+-sqrt(13))/2# Thus the critical points for #f(x)# are #{-1/2, (-1+sqrt(13))/2, (-1-sqrt(13))/2}#

Scarlett Allison · Answer

undefined To find the critical points of $ f(x) = \frac{1}{\sqrt{x+x^2-3}} $, we need to first find the derivative of $ f(x) $ and then solve for points where the derivative is either zero or undefined.

First, let's find the derivative of $ f(x) $:

$$ f(x) = \frac{1}{\sqrt{x+x^2-3}} $$
$$ f'(x) = \frac{d}{dx} \left( \frac{1}{\sqrt{x+x^2-3}} \right) $$

To differentiate $ \frac{1}{\sqrt{x+x^2-3}} $, we can use the chain rule:

$$ f'(x) = -\frac{1}{2} \cdot (x+x^2-3)^{-3/2} \cdot (1+2x) $$

To find critical points, we need to find where the derivative is zero or undefined. The derivative is undefined when the denominator $ (x+x^2-3) $ becomes zero, but this does not occur in the domain of $ f(x) $ since $ x+x^2-3 > 0 $ for all $ x $. So, the only critical points occur when the numerator $ (1+2x) $ equals zero:

$$ 1+2x = 0 $$
$$ x = -\frac{1}{2} $$

Therefore, the critical point of $ f(x) $ is $ x = -\frac{1}{2} $.