(When entering numerical values, if necessary round to two decimal places.) Consider the equation and the interval given below. In(x+1)x³= 0, (0.5,1) Part a) Use the Intermediate Value Theorem to show that there is a root to the given equation in the indicated interval. f(x) = ln(x + 1) - 2³ is defined continuous positive negative O O O on the closed interval [0.5, 1]. Moreover, f(0.5) = and f(1) = Therefore, f(0.5) < f(1) f(1) <f(0.5) Thus, f(0.5) f(1). Consequently, the conditions of the IVT are met. And, since y = 0 is between f(0.5) and f(1), the IVT guarantees the existence of x = c in the interval (0.5, 1) such that f(c) = 0. Therefore, there is a root to the equation In(x + 1)-³ = 0 in the interval (0.5, 1). Part b) Use your graphing calculator to find an interval of length 0.01 that contains a root. Enter your answer as an open interval: Question Help: Video

(When entering numerical values, if necessary round to two decimal

Question

(When entering numerical values, if necessary round to two decimal places.) Consider the equation and the interval given below. In(x+1)x³= 0, (0.5,1) Part a) Use the Intermediate Value Theorem to show that there is a root to the given equation in the indicated interval. f(x) = ln(x + 1) - 2³ is defined continuous positive negative O O O on the closed interval [0.5, 1]. Moreover, f(0.5) = and f(1) = Therefore, f(0.5) < f(1) f(1)