Prove, using methods taught in this course, that the equation has a solution on the given interval. Your work should be legible, and all your logic should be clear and justified. \( \quad-3 x^{2}-1=-2 e^{x}+3 \sqrt{x} \) on the interval \( (1,4) \) Choose File No file chosen

To prove that the equation \[ -3x^2 - 1 = -2e^x + 3\sqrt{x} \] has a solution on the interval \( (1, 4) \), we can rewrite the equation in the form of a function \( f(x) \): \[ f(x) = -3x^2 - 1 + 2e^x - 3\sqrt{x} \] We want to show that there exists at least one \( c \in (1, 4) \) such that \( f(c) = 0 \). To do this, we will use the Intermediate Value Theorem (IVT), which states that if \( f \) is continuous on the interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \in (a, b) \) such that \( f(c) = 0 \). ### Step 1: Check Continuity The function \( f(x) \) is composed of polynomial, exponential, and square root functions, all of which are continuous on their respective domains. Since \( \sqrt{x} \) is defined for \( x \geq 0 \) and all components of \( f(x) \) are continuous for \( x > 0 \), we conclude that \( f(x) \) is continuous on the interval \( (1, 4) \). ### Step 2: Evaluate \( f(x) \) at the endpoints of the interval Now we will evaluate \( f(x) \) at the endpoints \( x = 1 \) and \( x = 4 \). 1. **Evaluate \( f(1) \)**: \[ f(1) = -3(1)^2 - 1 + 2e^1 - 3\sqrt{1} \] \[ = -3 - 1 + 2e - 3 \] \[ = 2e - 7 \] Using \( e \approx 2.718 \): \[ f(1) \approx 2(2.718) - 7 \approx 5.436 - 7 \approx -1.564 \] Thus, \( f(1) < 0 \). 2. **Evaluate \( f(4) \)**: \[ f(4) = -3(4)^2 - 1 + 2e^4 - 3\sqrt{4} \] \[ = -48 - 1 + 2e^4 - 6 \] \[ = 2e^4 - 55 \] Using \( e^4 \approx 54.598 \): \[ f(4) \approx 2(54.598) - 55 \approx 109.196 - 55 \approx 54.196 \] Thus, \( f(4) > 0 \). ### Step 3: Apply the Intermediate Value Theorem We have found: - \( f(1) < 0 \) - \( f(4) > 0 \) Since \( f(x) \) is continuous on \( (1, 4) \) and \( f(1) < 0 \) while \( f(4) > 0 \), by the Intermediate Value Theorem, there exists at least one \( c \in (1, 4) \) such that \[ f(c) = 0. \] ### Conclusion Thus, we have proven that the equation \[ -3x^2 - 1 = -2e^x + 3\sqrt{x} \] has at least one solution in the interval \( (1, 4) \).