How do you find the critical points for #f(x) = 2x^(2/3)  5x^(4/3)# and the local max and min?
See the answer below:
Thanks Jim H for reminding me of the point of minimum at f (x) = 0.
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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.

Find the derivative ( f'(x) ): [ f'(x) = \frac{d}{dx} (2x^{2/3}  5x^{4/3}) ]

Compute the derivative: [ f'(x) = \frac{4}{3}x^{1/3}  \frac{20}{3}x^{1/3} ]

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 ]

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.
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