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

Ask by Potter Lambert. in the United States
Mar 12,2025

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There is at least one solution to the equation \( e^{-x} = 4 + x^3 \) in the interval \([-2, -1]\).

Solución

1. Define the function \[ f(x)=e^{-x}-x^3-4. \] This function is continuous on \([-2,-1]\) since it is composed of continuous functions. 2. Compute the value at \( x=-2 \): \[ f(-2)=e^{-(-2)}-(-2)^3-4=e^{2}-(-8)-4=e^{2}+8-4=e^{2}+4. \] Since \( e^{2}>0 \), we have \( f(-2)>0 \). 3. Compute the value at \( x=-1 \): \[ f(-1)=e^{-(-1)}-(-1)^3-4=e^{1}-(-1)-4=e+1-4=e-3. \] Since \( e\approx2.718 \), it follows that \( f(-1)<0 \). 4. Since \( f \) is continuous on \([-2,-1] \) and \[ f(-2)>0\quad\text{and}\quad f(-1)<0, \] by the Intermediate Value Theorem there exists a number \( c \) in the interval \([-2,-1]\) such that \[ f(c)=0. \] 5. Therefore, the equation \[ e^{-x}=4+x^3 \] has at least one solution in the interval \([-2,-1]\).

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The Deep Dive

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!

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