Let f(x) = 3x2 / 3 - x2. Find all critical points of f(x). Classify these critic

Question

Let f(x) = 3x2 / 3 - x2. Find all critical points of f(x). Classify these critical points into local minimum using the first derivative test.

Explanation / Answer

The first derivative is f'(x) = 2x^(-1/3) - 2x, which I'm sure you know how to find. To find the critical points, we must set the first derivative equal to 0 and solve for x: 2x^(-1/3) - 2x = 0 => 2x^(-1/3) = 2x => x^(-1/3) = x => 1 = x^(4/3) So the solutions are x = -1 and 1. (PART A) Now we plug in a point less than -8 into f to find its slope. In this case, I'll use -2: f'(-2) = 2(-8)^(-1/3) - 2(-8) = 2(-1/2) + 16 = -1 + 16 = 15 > 0 So we know that the function is increasing from (-infinity, -1). Thus clearly -1 is a local max because the function increases from the negative until it gets to -1, then it changes direction. Now we choose a point after 1. In this case I'll choose 8: f'(8) = 2(8)^(-1/3) -2(8) = 2(1/2) - 16 = 1 - 16 -15 < 0 So we know that the function is decreasing from (1, infinity). Thus 1 is also a local max.

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