Given \[ f(t)=\int_{0}^{t} \frac{x^{2}+12 x+20}{1+\cos ^{2}(x)} d x \] At what value of \( t \) does the local max of \( f(t) \) occur? \( t= \)

To find the value of \( t \) at which the local maximum of \( f(t) \) occurs, we need to analyze the critical points of the function. First, we find \( f'(t) \) using the Fundamental Theorem of Calculus, which states that if \( f(t) = \int_{a}^{t} g(x) \, dx \), then \( f'(t) = g(t) \). Thus, \[ f'(t) = \frac{t^{2}+12t+20}{1+\cos^{2}(t)}. \] To find the critical points where local maxima can occur, we set \( f'(t) = 0 \): \[ \frac{t^{2}+12t+20}{1+\cos^{2}(t)} = 0. \] Since \( 1 + \cos^2(t) \) is always positive, we need to solve: \[ t^{2}+12t+20 = 0. \] Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 12, c = 20 \): \[ t = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 80}}{2} = \frac{-12 \pm \sqrt{64}}{2} = \frac{-12 \pm 8}{2}. \] Calculating the two possible values: \[ t = \frac{-12 + 8}{2} = \frac{-4}{2} = -2, \] \[ t = \frac{-12 - 8}{2} = \frac{-20}{2} = -10. \] Both critical points are negative values. To determine which one corresponds to a local maximum, we can use the second derivative test or analyze the behavior around these points. Since \( f'(t) \) represents a quadratic function that opens upwards (as the coefficient of \( t^2 \) is positive), the vertex of this parabola occurs at the point between \( -2 \) and \( -10 \). Thus, we analyze behavior: - For \( t < -10 \), \( f'(t) > 0 \). - For \( -10 < t < -2 \), \( f'(t) < 0 \). - For \( t > -2 \), \( f'(t) > 0 \). Hence, \( t = -10 \) corresponds to a local maximum since the function is increasing to that point and decreasing afterward. Thus, the value of \( t \) where the local maximum of \( f(t) \) occurs is: \[ t = -10. \]