How do you find the critical numbers of #1/(t^2+3)#? Calculus Graphing with the First Derivative Identifying Stationary Points (Critical Points) for a Function 1 Answer Jim H Dec 8, 2017 The on;y critical number is #0#. Explanation: #f(t) = 1/(t^2+3)# #f'(t) = (-2t)/(t^2+3)^2# #f'(t)# is never undefined and is #0# only at #t=0# Answer link Related questions How do you find the stationary points of a curve? How do you find the stationary points of a function? How many stationary points can a cubic function have? How do you find the stationary points of the function #y=x^2+6x+1#? How do you find the stationary points of the function #y=cos(x)#? How do I find all the critical points of #f(x)=(x-1)^2#? Let #h(x) = e^(-x) + kx#, where #k# is any constant. For what value(s) of #k# does #h# have... How do you find the critical points for #f(x)=8x^3+2x^2-5x+3#? How do you find values of k for which there are no critical points if #h(x)=e^(-x)+kx# where k... How do you determine critical points for any polynomial? See all questions in Identifying Stationary Points (Critical Points) for a Function Impact of this question 1316 views around the world You can reuse this answer Creative Commons License