The critical points of a function f(x) are at the points where f'(x) = 0 or f'(x) does not exist.

For H(t) = t^(3/4) - 5*t^(1/4)

H'(t) = (3/4)*t^(-1/4) - 5*(1/4)*t^(-3/4)

H'(t) exists for all values of t except t = 0

Solving H'(t) = 0

=> (3/4)*t^(-1/4) - 5*(1/4)*t^(-3/4)...

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The critical points of a function f(x) are at the points where f'(x) = 0 or f'(x) does not exist.

For H(t) = t^(3/4) - 5*t^(1/4)

H'(t) = (3/4)*t^(-1/4) - 5*(1/4)*t^(-3/4)

H'(t) exists for all values of t except t = 0

Solving H'(t) = 0

=> (3/4)*t^(-1/4) - 5*(1/4)*t^(-3/4) = 0

=> 3*t^(-1/4) - 5*t^(-3/4) = 0

=> 3*t^(-1/4) = 5*t^(-3/4)

=> 3*t = 5*t^(-1/2)

=> 9t^2 = 25t

=> 9t = 25

=> t = 25/9

**The critical points are at t = 0 and t = 25/9**