Question
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

Upstudy AI Solution

Tutor-Verified Answer

Answer

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

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

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!

Related Questions

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 \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
\[ y=x^{3}, \quad 0 \leq x \leq 2 \] 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 \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]

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 \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy