How do you find the critical points for #f(x) = 2x^(2/3) - 5x^(4/3)# and the local max and min?

Question

Evelyn Agard · Accepted Answer

See the answer below:
Thanks Jim H for reminding me of the point of minimum at f (x) = 0.

Cameron Alexander · Answer

undefined To find the critical points of $ f(x) = 2x^{2/3} - 5x^{4/3} $, we first find the derivative $ f'(x) $ and set it equal to zero to locate the critical points. Then, we determine whether each critical point corresponds to a local maximum, local minimum, or neither by using the first derivative test. 1. Find the derivative $ f'(x) $: $$ f'(x) = \frac{d}{dx} (2x^{2/3} - 5x^{4/3}) $$ 2. Compute the derivative: $$ f'(x) = \frac{4}{3}x^{-1/3} - \frac{20}{3}x^{1/3} $$ 3. Set $ f'(x) $ equal to zero and solve for $ x $: $$ \frac{4}{3}x^{-1/3} - \frac{20}{3}x^{1/3} = 0 $$ $$ \frac{4}{3}x^{-1/3} = \frac{20}{3}x^{1/3} $$ $$ 4x = 20 $$ $$ x = 5 $$ 4. Determine the nature of the critical point: Plug in values around $ x = 5 $ into $ f'(x) $ to see the behavior of the function. For $ x < 5 $, $ f'(x) < 0 $, so the function is decreasing. For $ x > 5 $, $ f'(x) > 0 $, so the function is increasing. Therefore, at $ x = 5 $, there is a local minimum. Thus, the critical point is $ (5, f(5)) $, and it corresponds to a local minimum.