Q:

# Find the x-coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method. f(x) = 5x4 β 2x3

Accepted Solution

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Answer:x = 0, local maximumx = Β±(2/3)β3, global minimaStep-by-step explanation:The first derivative is ... Β  f'(x) = 30x^5 -40x^3 = 10x^3(3x^2 -4)This will have zeros (critical points) at x=0 and x=Β±β(4/3).*We don't need the second derivative to tell the nature of these critical points. Since the degree is even, the function is symmetrical about x=0. Since the leading coefficient is positive, it generally has a U-shape. This means the "outer" critical points will be minima, and the central one will be a local maximum.__However, since we're asked to use the 2nd derivative test first, we find the 2nd derivative to be ... Β  f''(x) = 150x^4 -120x^2 = 30x^2(5x^2 -4)For x=0, f''(0) = 0 -- as we expect for a function with a high multiplicity of the root at that point. For x either side of zero, both the function and the second derivative are negative, indicating downward concavity. That is, x = 0 is a local maximum.For xΒ² = 4/3, the second derivative is positive, indicating upward concavity. At x = Β±β(4/3), we have local minima._____* The "simplified" equivalent to β(4/3) is (2/3)β3.