Q1- Show by intermediate value theorem that the equation \( e^{-x}=4+x^{3} \) is solvable on the interval \( [-2,-1] \) Hint: \( f(x)=e^{-x}-x^{3}-4=0 \)

To show that the equation \( e^{-x} = 4 + x^3 \) is solvable on the interval \( [-2, -1] \) using the Intermediate Value Theorem, we can define the function \( f(x) = e^{-x} - x^3 - 4 \). We need to evaluate \( f(-2) \) and \( f(-1) \): First, calculating \( f(-2) \): \[ f(-2) = e^{2} - (-2)^3 - 4 = e^{2} + 8 - 4 = e^{2} + 4 \approx 7.39 > 0. \] Next, calculating \( f(-1) \): \[ f(-1) = e^{1} - (-1)^3 - 4 = e + 1 - 4 \approx 2.72 - 4 = -1.28 < 0. \] Since \( f(-2) > 0 \) and \( f(-1) < 0 \), and \( f(x) \) is continuous (being a combination of exponential and polynomial functions), it follows by the Intermediate Value Theorem that there exists at least one \( c \) in the interval \( (-2, -1) \) such that \( f(c) = 0 \). Thus, the equation \( e^{-x} = 4 + x^3 \) has at least one solution in the interval \( [-2, -1] \). So, there you have it—an engaging evidence that this equation is indeed solvable in the specified range! It's a delightful dance of numbers and continuity. And as for fun, who knew mathematics could lead to a treasure hunt in such intervals? Keep those calculators handy; you never know when a wandering solution might pop up from thin air!