How do you find the critical numbers for #h(t) = (t^(3/4)) − (9*t^(1/4))# to determine the maximum and minimum?

1 Answer

Answer:

#(9, -2*9^0.75)#

Explanation:

Differentiate #h(t)# with respect to t.

#h(t)=t^(3/4)-9*t^(1/4)#

#h'(t)=(3/4)*t^(3/4-1)-9(1/4)t^(1/4-1)#

#h'(t)=(3/4)*t^(-1/4)-(9/4)t^(-3/4)#

After finding the derivative #h'(t)#
Set #h'(t)=0#
#(3/4)*t^(-1/4)-(9/4)t^(-3/4)=0#
Multiply both sides by #t^(3/4)#
#t^(3/4)((3/4)*t^(-1/4)-(9/4)t^(-3/4))=0*t^(3/4)#

#(3/4)*t^(1/2)-(9/4)t^0=0#
#(3/4)*t^(1/2)-9/4=0#
#(3/4)*t^(1/2)=9/4#

#t^(1/2)=3#

#t=9#

Solve now for #f(9)# using the given function #h(t)=t^(3/4)-9*t^(1/4)#

#h(9)=9^(3/4)-9*9^(1/4)#
#h(9)=9^(3/4)-9^(5/4)#
#h(9)=9^(3/4)(1-9^(1/2))#
#h(9)=9^(3/4)(1-3)#
#h(9)=9^(3/4)(-2)#
#h(9)=-2*9^(3/4)#

The point #(9, -2*9^(3/4))# is a minimum

graph{y=x^(3/4)-9*x^(1/4) [-40, 40, -20, 10]}

God bless....I hope the explanation is useful.