How do you find the critical numbers for h(t) = (t^(3/4)) − (9*t^(1/4))h(t)=(t34)(9t14) to determine the maximum and minimum?

1 Answer

(9, -2*9^0.75)(9,290.75)

Explanation:

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

h(t)=t^(3/4)-9*t^(1/4)h(t)=t349t14

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.