What are the absolute extrema of #f(x)=9x^(1/3)-3x in[0,5]#?

Question

Elijah Abram · Accepted Answer

The absolute maximum of #f(x)# is #f(1)=6# and the absolute minimum is #f(0)=0#.

To find the absolute extrema of a function, we need to find its critical points. These are the points of a function where its derivative is either zero or does not exist. The derivative of the function is #f'(x)=3x^(-2/3)-3#. This function (the derivative) exists everywhere. Let's find where it is zero: #0=3x^(-2/3)-3rarr3=3x^(-2/3)rarrx^(-2/3)=1rarrx=1# We also have to consider the endpoints of the function when looking for absolute extrema: so the three possibilities for extrema are #f(1), f(0)# and # f(5)#. Calculating these, we find that #f(1)=6, f(0)=0,# and #f(5)=9root(3)(5)-15~~0.3#, so #f(0)=0# is the minimum and #f(1)=6# is the max.

Elijah Abram · Answer

undefined To find the absolute extrema of $ f(x) = 9x^{1/3} - 3x $ in the interval $[0, 5]$, we first evaluate $ f(x) $ at the critical points and endpoints within the interval, and then compare their values.

1. Find the critical points by setting the derivative of $ f(x) $ equal to zero and solving for $ x $:
   $$ f'(x) = \frac{d}{dx}(9x^{1/3} - 3x) = 3x^{-2/3} - 3 = 0 $$
   Solving this equation, we find $ x = 1 $ as the critical point.

2. Evaluate $ f(x) $ at the critical point and endpoints:
   $$ f(0) = 9(0)^{1/3} - 3(0) = 0 $$
   $$ f(1) = 9(1)^{1/3} - 3(1) = 9 - 3 = 6 $$
   $$ f(5) = 9(5)^{1/3} - 3(5) = 9(125^{1/3}) - 15 = 9(5) - 15 = 45 - 15 = 30 $$

3. Compare the values obtained:
   - At $ x = 0 $, $ f(x) = 0 $
   - At $ x = 1 $, $ f(x) = 6 $
   - At $ x = 5 $, $ f(x) = 30 $

4. Therefore, the absolute minimum value of $ f(x) $ in the interval $[0, 5]$ is $ 0 $ which occurs at $ x = 0 $, and the absolute maximum value is $ 30 $ which occurs at $ x = 5 $.