Find the value of x at which the following function has a local minimum: f\left(x\right)=\frac{x}{\left(x^2-5\right)^{\frac{1}{3}}}

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The problem asks for the value of x where the function f(x) = x / (x^2 - 5)^(1/3) has a local minimum. First, identify the domain: (x^2 - 5)^(1/3) is a real cube root, which is defined for all real x, but the denominator cannot be zero. x^2 - 5 = 0 at x = ±√5, so the function is undefined at x = ±√5 and the domain splits there. Next, compute the derivative to locate critical points. Write f(x) as x · (x^2 - 5)^(-1/3) and apply the product rule: f'(x) = (x^2 - 5)^(-1/3) + x · (-1/3)(x^2 - 5)^(-4/3) · (2x). This simplifies to combining like terms with a common factor: f'(x) = (1/3)(x^2 - 15) · (x^2 - 5)^(-4/3). Critical points occur when f'(x) = 0 or is undefined. Since (x^2 - 5)^(-4/3) is positive wherever it is defined (and undefined only at x = ±√5, which are not in the domain), the sign and zeros come from the (x^2 - 15) factor. Thus the derivative vanishes when x^2 - 15 = 0, giving x = ±√15 ≈ ±3.873. Now determine which critical points correspond to local minima or maxima by sign-testing f'(x) around the critical values, keeping in mind the derivative factor (x^2 - 5)^(-4/3) stays positive on each interval of the domain: - For |x| > √15, x^2 - 15 > 0, so f'(x) > 0. - For √5 < |x| < √15, x^2 - 15 < 0, so f'(x) < 0. - For |x| < √5, we still have x^2 - 15 < 0, hence f'(x) < 0 as well (since the other factor is positive). From this sign pattern, at x = -√15 the derivative changes from positive to negative, indicating a local maximum there. At x = √15 the derivative changes from negative to positive, indicating a local minimum there. Therefore, the local minimum occurs at the positive critical point x = √15. The value is x ≈ 3.873. For completeness, note how the function behaves near the asymptotes x = ±√5: as x approaches ±√5 from either side, the denominator tends to zero and the function diverges, which is consistent with the domain gaps and the presence of separate branches. In summary, the candidate for a local minimum is at x = √15, with an approximate numerical value around 3.873. The negative counterpart x = -√15 corresponds to a local maximum, not a minimum.

ECON6003/ECON6703 Mathematical Methods of Econ Analysis Assignment 1

Find the value of x at which the following function has a local minimum: f\left(x\right)=\frac{x}{\left(x^2-5\right)^{\frac{1}{3}}}

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