Find the critical numbers of the function : h(t) = t^3/4 - 2t^1/4

Question

Find the critical numbers of the function : h(t) = t^3/4 - 2t^1/4 

 

I know I first need to find the derivative:

 

3/4 t^-1/4 - 1/2 t^-3/4

 

I know I need to set it all equal to zero, but I'm not sure how to solve for t from there.

 

p281, #37

ANSWER: 0, 4/9

Answer 1

h(t) = t^3/4 - 2t^1/4

 

The critical numbers are values of t that are in the domain of h(t) and for which h'(t) = 0 or h'(t) is undefined.  

 

h'(t) = (3/4)t^-1/4 - (1/2)t^-3/4  = 3/(4t^1/4) - 1/(2t^3/4)

 

                                         = (3t^1/2 - 2)/(4t^3/4)

 

h'(t) = 0 when 3t^1/2 = 2

 

                      t^1/2 = 2/3

 

                      t = 4/9

 

h(0) is defined but h'(0) is not defined, so 0 is also a critical number

 

The critical numbers are 4/9 and 0.