How do you find the critical numbers for #root3((x^2-x))# to determine the maximum and minimum?

Question

Emma Aldridge · Accepted Answer

The critical number is #x=1/2#.

First we take the derivative using the chain rule, to make things easier for us we rewrite the problem using powers. #(x^2-x)^(1/3)# Now we apply the chain rule we take the derivative of the outside and multiple it by the derivative of the inside. It's important that you know the power rule. #d/dx# #=# #1/3(x^2-x)^(-2/3)(2x-1)# Now we rewrite it: #d/dx# #=# #(2x-1)/(3root(3)((x^2-x)^2)# Set it equal to zero and solve: #(2x-1)/(3root(3)((x^2-x)^2)# #=0# #cancelcolor(green)(3root(3)((x^2-x)^2))/1**(2x-1)/cancelcolor(green)(3root(3)((x^2-x)^2)# #=0# #(3root(3)((x^2-x)^2))# We are left with: #2x-1=0# Solve: #x=1/2# By looking at the graph you can see that #x=1/2# is a critical number. graph{(x^2-x)^(1/3) [-1.912, 4.248, -1.023, 2.057]}

Elijah Abram · Answer

undefined To find the critical numbers for the function $ f(x) = \sqrt[3]{x^2 - x} $, and subsequently determine the maximum and minimum, we need to follow these steps:

1. Find the derivative of the function $ f(x) $.
2. Set the derivative equal to zero and solve for $ x $.
3. Evaluate the second derivative to determine whether the critical points correspond to a maximum, minimum, or neither.

Let's proceed with these steps.

Luke Alvarez · Answer

undefined To find the critical numbers of the function $ \sqrt{3(x^2 - x)} $, we first need to find its derivative and then determine where the derivative is equal to zero or undefined.

Let's denote the function as $ f(x) = \sqrt{3(x^2 - x)} $. 
To find the derivative, $ f'(x) $, we can use the chain rule:

$$ f'(x) = \frac{d}{dx} \sqrt{3(x^2 - x)} $$

$$ = \frac{1}{2\sqrt{3(x^2 - x)}} \cdot \frac{d}{dx} (3(x^2 - x)) $$

$$ = \frac{1}{2\sqrt{3(x^2 - x)}} \cdot (6x - 3) $$

Now, we need to find where $ f'(x) = 0 $ or $ f'(x) $ is undefined.

Setting $ f'(x) = 0 $ and solving for $ x $:

$$ \frac{1}{2\sqrt{3(x^2 - x)}} \cdot (6x - 3) = 0 $$

$$ 6x - 3 = 0 $$

$$ x = \frac{1}{2} $$

To check for points where $ f'(x) $ is undefined, we need to look for any values of $ x $ that make the denominator of $ f'(x) $ equal to zero. However, in this case, the function is defined for all real numbers, so there are no additional critical points from the denominator being zero.

So, the only critical number is $ x = \frac{1}{2} $.

To determine whether this critical point corresponds to a maximum or minimum, we can use the second derivative test or analyze the behavior of the function around this point. However, as the function $ f(x) = \sqrt{3(x^2 - x)} $ is always increasing for $ x > \frac{1}{2} $ and always decreasing for $ x < \frac{1}{2} $, the critical point $ x = \frac{1}{2} $ corresponds to a minimum value of the function.