How do you find all critical points (if any) if #k(t)=1/sqrt(t^2 +1)#?

Question

Emily Ammerman · Accepted Answer

Hello, Critical points of the function #k# are the numbers #t# which are solutions of #k'(t)=0#. Here, #k(t) = (t^2+1)^{-1/2}#, therefore #k'(t) = -\frac{1}{2}(t^2+1)^{-3/2}	imes (2t)#. So, #k'(t) = 0 iff t=0#. The only critical point of #k# is #t=0#.

Paisley Johnson · Answer

undefined To find the critical points of the function $ k(t) = \frac{1}{\sqrt{t^2 + 1}} $, we need to find the values of $ t $ where the derivative of $ k(t) $ is equal to zero or is undefined.

1. Find the derivative of $ k(t) $ with respect to $ t $:
   $$ k'(t) = -\frac{t}{(t^2 + 1)^{3/2}} $$

2. Set the derivative equal to zero and solve for $ t $ to find the points where the derivative is zero:
   $$ -\frac{t}{(t^2 + 1)^{3/2}} = 0 $$
   Since the numerator is zero, the derivative is zero when $ t = 0 $.

3. Determine if there are any points where the derivative is undefined. The derivative is undefined when the denominator of the derivative expression is zero, but $ (t^2 + 1)^{3/2} $ is never zero for real $ t $. So, there are no points where the derivative is undefined.

Therefore, the only critical point of $ k(t) $ is at $ t = 0 $.