Pregunta
upstudy study bank question image url

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

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

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 \([-2,-1]\).

Respondido por UpStudy AI y revisado por un tutor profesional

error msg
Explicar
Simplifique esta solución

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!

Latest Calculus Questions

\[ y=x^{3}, \quad 0 \leq x \leq 2 \] Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interv \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note t as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius and the radical measures the arc length that is the width of a band, \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying, \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9} \sqrt[9]{ } x^{4} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 14 \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad