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

The Intermediate Value Theorem (IVT) states that if a continuous function changes sign over an interval, then there is at least one root in that interval. Let's evaluate the function \( f(x) = e^{-x} - x^3 - 4 \) at the endpoints of the interval \( [-2, -1] \). Calculating \( f(-2) \): \[ f(-2) = e^2 - (-2)^3 - 4 = e^2 + 8 - 4 = e^2 + 4 \] Since \( e^2 \approx 7.39 \), we get \( f(-2) \approx 11.39 > 0 \). Now evaluating \( f(-1) \): \[ f(-1) = e^1 - (-1)^3 - 4 = e + 1 - 4 = e - 3 \] With \( e \approx 2.71 \), we find \( f(-1) \approx -0.29 < 0 \). Thus, we have \( f(-2) > 0 \) and \( f(-1) < 0 \), confirming that \( f(x) \) changes sign over the interval \( [-2, -1] \). By the IVT, there is at least one root in that interval. To visualize how this theorem works in action, picture a rollercoaster! If one car starts high up on the left (positive height) and by the end reaches a low point on the right (negative height), there has to be a point in the middle where it dips below the ground—just like our function crossing the x-axis! As a fun fact, the Intermediate Value Theorem applies not only to polynomials but also to any continuous functions—like trigonometric and exponential functions. So next time you're stuck in a math situation, check if the function you’re dealing with is continuous and have a sign change; you've got IVT on your side!