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Step 2: Set the first derivative equal to 0 to solve for critical points: $\\frac{1}{x^{\\frac{2}{3}}} - 2 = 0 \\implies \\frac{1}{x^{\\frac{2}{3}}} = 2 \\implies x^{\\frac{2}{3}} = \\frac{1}{2}$. Cube both sides: $x^2 = \\frac{1}{8}$, so $x = \\pm\\frac{1}{2\\sqrt{2}} = \\pm\\frac{\\sqrt{2}}{4} \\approx \\pm0.353$. Only $x = \\frac{1}{8}$ (since $\\left(\\frac{1}{8}\\right)^{\\frac{2}{3}} = \\frac{1}{4}$? Correction: Solve $x^{\\frac{2}{3}}=1/2$: square both sides? No, raise to 3/2 power: $x=(1/2)^{3/2}=1/(2\\sqrt{2})=\\sqrt{2}/4≈0.353$, but $\\sqrt{2}/4$ squared is 2/16=1/8, so $x=1/8$ is when $x^{2/3}=1/4$, no—wait, correct critical point: $1/x^{2/3}=2 \\implies x^{2/3}=1/2 \\implies x=(1/2)^{3/2}=1/(2\\sqrt{2})=\\sqrt{2}/4≈0.353$, but let's test values. Wait, evaluate the function at critical points and endpoints: $x=0$: $y=0$; $x=1$: $y=3-2=1$; critical point $x=(1/2)^{3/2}=1/(2\\sqrt{2})≈0.353$, $y=3*(1/(2\\sqrt{2}))^{1/3}-2*(1/(2\\sqrt{2}))=3*(1/2^{1/2}) - 2*(1/(2\\sqrt{2}))= 3/\\sqrt{2} - 1/\\sqrt{2}=2/\\sqrt{2}=\\sqrt{2}≈1.414$. Wait, no, check derivative sign: for $x<(1/2)^{3/2}$, $y'>0$; for $x>(1/2)^{3/2}$, $y'<0$? No, wait $x^{-2/3}$ is always positive, so when $x<(1/2)^{3/2}$, $1/x^{2/3}>2$, so $y'>0$; when $x>(1/2)^{3/2}$, $1/x^{2/3}<2$, so $y'<0$. Wait, but that would mean the critical point is a maximum. Then check endpoints: $x=0$ gives $y=0$, $x=1$ gives $y=1$, so the minimum is at $x=0$? No, wait, wait derivative calculation: $d/dx [3x^{1/3}]=3*(1/3)x^{-2/3}=x^{-2/3}$, correct. $d/dx[-2x]=-2$, correct. Wait, maybe I made a mistake in critical point: $x^{-2/3}=2 \\implies x^{2/3}=1/2 \\implies x=(1/2)^{3/2}=1/(2\\sqrt{2})≈0.353$, which is in [0,1]. Now evaluate the function at $x=0$: $y=0$; $x=1/(2\\sqrt{2})$: $y=3*(1/(2\\sqrt{2}))^{1/3} - 2*(1/(2\\sqrt{2}))$. $(1/(2\\sqrt{2}))^{1/3}=(1/2^{3/2})^{1/3}=1/2^{1/2}=1/\\sqrt{2}$. So $y=3*(1/\\sqrt{2}) - 2*(1/(2\\sqrt{2}))= 3/\\sqrt{2} - 1/\\sqrt{2}=2/\\sqrt{2}=\\sqrt{2}≈1.414$. $x=1$: $y=3-2=1$. Wait, but the minimum is at $x=0$? But that's option A. Wait no, wait maybe I messed up the derivative sign: for $x>0$, $x^{-2/2}=1/x^{2/3}$, which is always positive. When $x=1/8$, $x^{2/3}=(1/8)^{2/3}=(1/2^3)^{2/3}=1/2^2=1/4$, so $1/x^{2/3}=4$, so $y'=4-2=2>0$. When $x=1/2$, $x^{2/3}=(1/2)^{2/3}≈0.63$, $1/x^{2/3}≈1.587<2$, so $y'=1.587-2≈-0.413<0$. Oh! Wait, I solved $x^{2/3}=1/2$ wrong: $x^{2/3}=1/2$ implies $x=(1/2)^{3/2}=1/(2\\sqrt{2})≈0.353$, which is between 1/8 and 1/2. So the function increases on [0, (1/2)^{3/2}] and decreases on [(1/2)^{3/2}, 1]. So the maximum is at (1/2)^{3/2}, and the minimum is at the endpoints. Evaluate y at x=0: 0; x=1: 1. So the minimum is at x=0? But that's option A. Wait no, wait the question says minimum value. Wait maybe I made a mistake in the derivative. Wait no, let's check the function: $y=3x^{1/3}-2x$. At x=1/8: $y=3*(1/8)^{1/3}-2*(1/8)=3*(1/2)-1/4=3/2-1/4=5/4=1.25$, which is greater than 0. At x=0, y=0, which is smaller. Wait, but why is option B 1/8? Oh wait, maybe I misread the function: is it $y=3x^{1/3}-2x$ or $y=3x^{3}-2x$? No, the question says $3x^{\\frac{1}{3}} - 2x$. Wait, maybe the question is asking for the local minimum? No, it says on the closed interval [0,1]. Wait, maybe I messed up the derivative sign: when x approaches 0 from the right, $x^{-2/3}$ approaches infinity, so y' approaches infinity, so the function is increasing at x=0. Then it increases until x=(1/2)^{3/2}, then decreases. So the function goes from 0 up to sqrt(2)≈1.414, then down to 1. So the minimum is at x=0, which is option A. But wait, maybe I made a mistake in critical point calculation. Wait, let's use the second derivative test: $y''= -\\frac{2}{3}x^{-\\frac{5}{3}}$, which is negative for all x>0, so the critical point is a local maximum. So the function has a local maximum at x=(1/2)^{3/2}, and the minimum on [0,1] is at x=0. But wait, the options include 1/8. Oh! Wait, maybe I solved $x^{\\frac{2}{3}}=1/2$ wrong: $x^{\\frac{2}{3}}=1/2$ => $x^2=(1/2)^3=1/8$ => $x=1/(2\\sqrt{2})=\\sqrt{2}/4≈0.353$, which is not 1/8. 1/8 is 0.125. Wait, maybe the question is $y=3x^{3}-2x$? No, the question says $3x^{\\frac{1}{3}} - 2x$. Wait, maybe I misread the interval: is it [0,1] or [0, infinity)? No, it's [0,1]. Wait, let's check the value at x=1/8: y=3*(1/8)^{1/3}-2*(1/8)=3*(1/2)-1/4=3/2-1/4=5/4=1.25, which is greater than 0. At x=0, y=0, which is smaller. So the minimum is at x=0, option A. But wait, maybe the question is asking for the local minimum? But on [0,1], there is no local minimum except the endpoint. Wait, maybe I made a mistake in the derivative: $d/dx [3x^{1/3}]=3*(1/3)x^{-2/3}=x^{-2/3}$, correct. $d/dx[-2x]=-2$, correct. So y'=x^{-2/3}-2. Set to 0: x^{-2/3}=2 => x^{2/3}=1/2 => x=(1/2)^{3/2}=1/(2\\sqrt{2})≈0.353, which is in [0,1]. So the function increases on [0, (1/2)^{3/2}] and decreases on [(1/2)^{3/2}, 1]. So the minimum is at x=0, which is option A. But why is option B 1/8? Maybe the question was written incorrectly? Or maybe I misread the function as $3x^{1/3}-2x$ instead of $3x^{3}-2x$? If it's $3x^3-2x$, then y'=9x^2-2, set to 0: x=sqrt(2/9)=sqrt(2)/3≈0.471, which is not 1/8. Wait, no. Wait, maybe the function is $y=3x^{1/2}-2x$? Then y'= (3/2)x^{-1/2}-2, set to 0: (3/2)x^{-1/2}=2 => x^{-1/2}=4/3 => x=9/16, which is not 1/8. Wait, maybe the question is asking for the maximum? Then the maximum is at x=(1/2)^{3/2}=sqrt(2)/4≈0.353, which is not an option. Wait, the options are (A)0, (B)1/8, (C)1/4, (D)1. Oh! Wait, wait a second: $x^{-2/3}=2$ => $1/x^{2/3}=2$ => $x^{2/3}=1/2$ => $(x^{1/3})^2=1/2$ => $x^{1/3}=1/\\sqrt{2}$ => $x=1/(\\sqrt{2})^3=1/(2\\sqrt{2})=\\sqrt{2}/4≈0.353$, which is 1/(2.828)≈0.353, which is not 1/8. But 1/8 is 0.125. Wait, maybe I made a mistake in the sign of the derivative: if y'=2 - x^{-2/3}, then set to 0: x^{-2/3}=2, same result. No. Wait, let's evaluate the function at all options: (A)x=0: y=0; (B)x=1/8: y=3*(1/8)^{1/3}-2*(1/8)=3*(1/2)-1/4=3/2-1/4=5/4=1.25; (C)x=1/4: y=3*(1/4)^{1/3}-2*(1/4)=3*(1/2^{2/3})-1/2≈3*(0.63)-0.5≈1.89-0.5=1.39; (D)x=1: y=3-2=1. So the smallest value is at x=0, which is option A. But maybe the question is asking for the local minimum? But there is no local minimum on (0,1), only a local maximum. So the answer must be (A)0. Wait, but maybe I misread the function: is it $y=3x^{1/3}-2x$ or $y=2x-3x^{1/3}$? If it's $y=2x-3x^{1/3}$, then y'=2 - x^{-2/3}, set to 0: x^{-2/3}=2, same critical point, and the function decreases on [0, (1/2)^{3/2}] and increases on [(1/2)^{3/2}, 1], so the minimum is at x=(1/2)^{3/2}, which is not an option. But the question says $y=3x^{1/3}-2x$. So the correct answer is (A)0. Wait, but maybe the question is from an old exam, and I made a mistake. Wait, let's check the second derivative: y''= -2/3 x^{-5/3}, which is negative for all x>0, so the function is concave down on (0,1), so the critical point is a local maximum, so the minimum must be at one of the endpoints. x=0 gives y=0, x=1 gives y=1, so the minimum is at x=0. So the correct answer is (A)0.
