4.1 Max and Mins

Question

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

h(t) = t^3/4 − 9t^1/4

t = 

Answer 1

This problem is asking you to find the critical numbers for the function, i.e. where the 1st derivative is zero or undefined. Note that the domain of the function is x ≥ 0

f'(x) = 3/(4x^1/4) - 9/(4x^3/4)

This is clearly undefined when x = 0 (division by zero).

Are there any other numbers where 1st derivative is equal to zero?

Setting equal to zero and multiplying every term by 4/3 leaves:

1/x^1/4 - 3/x^3/4 = 0

Now multiply every term by x^3/4 (remember x cannot equal zero)..

x^1/2 - 3 = 0

√x = 3

x = 9

So, 0 and 9 are the critical numbers.

The first derivative is negative on 0<x<9 and positive on 9 <x<≈, so there is a relative min at (9,f(9)), which turns out to be an absolute min. The function goes to infinity as x -> ≈, so there is no absolute max unless the domain is restricted to some close interval.

Note that there is a vertical tangent line at (0,0).

desmos.com/calculator/1r1foqfk3h

Answer 2

Determine f'(t) and set equal to zero.

Max is at t = 0