Given the function \( f(x)=\int_{2}^{x}\left(t^{4}-4 t^{2}\right) d t \), determine all \( x \)-values, if any, for which \( f \) has a relative minimum.
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To determine the \( x \)-values for which the function \( f(x) = \int_{2}^{x}(t^{4}-4t^{2}) \, dt \) has a relative minimum, we first need to find the derivative \( f'(x) \), which is given by the Fundamental Theorem of Calculus: \[ f'(x) = x^4 - 4x^2. \] Next, we set the derivative equal to zero to find the critical points: \[ x^4 - 4x^2 = 0. \] Factoring out \( x^2 \): \[ x^2(x^2 - 4) = 0. \] This gives us: \[ x^2 = 0 \quad \text{or} \quad x^2 - 4 = 0. \] Thus, we have two critical points: \[ x = 0 \quad \text{and} \quad x = \pm 2. \] To determine whether these critical points correspond to relative minimums or maximums, we can use the second derivative test. The second derivative is: \[ f''(x) = 4x^3 - 8x. \] Evaluating \( f''(x) \) at the critical points: 1. For \( x = 0 \): \[ f''(0) = 4(0)^3 - 8(0) = 0. \] (So we cannot conclude about this point from the second derivative test.) 2. For \( x = 2 \): \[ f''(2) = 4(2)^3 - 8(2) = 32 - 16 = 16 \quad (> 0) \quad \text{(indicating a local minimum)}. \] 3. For \( x = -2 \): \[ f''(-2) = 4(-2)^3 - 8(-2) = -32 + 16 = -16 \quad (< 0) \quad \text{(indicating a local maximum)}. \] Thus, the only critical point where \( f(x) \) has a relative minimum is at \( x = 2 \).
