Find the critical numbers of the function. h(t) = t3/4 − 4t1/4

Answer 1

Take the derivative and 1) find where the derivative equals zero, 2) find where the derivative does not exist:

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

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

h'(t) = 3/(4t^1/4) - 1/[(t³)^1/4]

We can see the x = 0 makes the derivative function undefined so x = 0 is a critical number.

3/(4t^1/4) - 1/[(t³)^1/4] = 0

3/(4t^1/4) = 1/[(t³)^1/4]

3[(t³)^1/4] = 1(4t^1/4)

3(t³)^1/4 = 4t^1/4

Raise both sides to the 4th power to get:

81t³ = 256t

81t³ - 256t = 0

t(81t² - 256)

t(9t + 16)(9t - 16)

t = 0, t = -16/9, t = 16/9

We already listed t = 0 as a critical number. Looking at the other results, we are going to have to throw out the x = -16/9 because the original function is a 4th root function which we are unable to do with a negative argument (at least for real numbers). So the critical numbers are x = 0 and x = 16/9