Math Problem Statement

Use the Intermediate Value Theorem to show that there is a solution to the given equation in the specified interval: i. [4 points] ln(x) = 4 − √ x, for x ∈ (4,9). 1 ii. [4 points] e x = 3 − 2x, for x ∈ (0,1)

Solution

The Intermediate Value Theorem (IVT) states that if a function f(x)f(x) is continuous on a closed interval [a,b][a, b], and if f(a)f(a) and f(b)f(b) have opposite signs, then there exists at least one c(a,b)c \in (a, b) such that f(c)=0f(c) = 0.

Part i: ln(x)=4x, for x(4,9)\ln(x) = 4 - \sqrt{x}, \text{ for } x \in (4,9)

We can rewrite the equation as a function:

f(x)=ln(x)(4x)f(x) = \ln(x) - (4 - \sqrt{x}) We aim to find if there exists c(4,9)c \in (4,9) such that f(c)=0f(c) = 0, by applying the IVT.

Step 1: Check Continuity

  • Both ln(x)\ln(x) and x\sqrt{x} are continuous for x(4,9)x \in (4,9).
  • Therefore, f(x)=ln(x)(4x)f(x) = \ln(x) - (4 - \sqrt{x}) is also continuous on (4,9)(4,9).

Step 2: Evaluate f(x)f(x) at the endpoints of the interval

  1. At x=4x = 4: f(4)=ln(4)(44)=ln(4)(42)=ln(4)2f(4) = \ln(4) - (4 - \sqrt{4}) = \ln(4) - (4 - 2) = \ln(4) - 2 Using ln(4)1.386\ln(4) \approx 1.386, we get: f(4)1.3862=0.614f(4) \approx 1.386 - 2 = -0.614

  2. At x=9x = 9: f(9)=ln(9)(49)=ln(9)(43)=ln(9)1f(9) = \ln(9) - (4 - \sqrt{9}) = \ln(9) - (4 - 3) = \ln(9) - 1 Using ln(9)2.197\ln(9) \approx 2.197, we get: f(9)2.1971=1.197f(9) \approx 2.197 - 1 = 1.197

Step 3: Apply the IVT

  • f(4)0.614f(4) \approx -0.614 and f(9)1.197f(9) \approx 1.197, so f(4)<0f(4) < 0 and f(9)>0f(9) > 0.
  • Since f(x)f(x) is continuous and changes sign between x=4x = 4 and x=9x = 9, the IVT guarantees that there is at least one solution c(4,9)c \in (4,9) such that f(c)=0f(c) = 0.

Conclusion for Part i:

There exists a solution to the equation ln(x)=4x\ln(x) = 4 - \sqrt{x} in the interval (4,9)(4,9).


Part ii: ex=32x, for x(0,1)e^x = 3 - 2x, \text{ for } x \in (0,1)

We rewrite the equation as a function:

g(x)=ex(32x)g(x) = e^x - (3 - 2x) We aim to show there is a solution c(0,1)c \in (0,1) such that g(c)=0g(c) = 0, using the IVT.

Step 1: Check Continuity

  • Both exe^x and 32x3 - 2x are continuous for all x(0,1)x \in (0,1).
  • Therefore, g(x)=ex(32x)g(x) = e^x - (3 - 2x) is continuous on (0,1)(0,1).

Step 2: Evaluate g(x)g(x) at the endpoints of the interval

  1. At x=0x = 0: g(0)=e0(32(0))=13=2g(0) = e^0 - (3 - 2(0)) = 1 - 3 = -2

  2. At x=1x = 1: g(1)=e1(32(1))=e(32)=e1g(1) = e^1 - (3 - 2(1)) = e - (3 - 2) = e - 1 Using e2.718e \approx 2.718, we get: g(1)2.7181=1.718g(1) \approx 2.718 - 1 = 1.718

Step 3: Apply the IVT

  • g(0)=2g(0) = -2 and g(1)1.718g(1) \approx 1.718, so g(0)<0g(0) < 0 and g(1)>0g(1) > 0.
  • Since g(x)g(x) is continuous and changes sign between x=0x = 0 and x=1x = 1, the IVT guarantees that there is at least one solution c(0,1)c \in (0,1) such that g(c)=0g(c) = 0.

Conclusion for Part ii:

There exists a solution to the equation ex=32xe^x = 3 - 2x in the interval (0,1)(0,1).


Do you want more details on any step, or further explanation?

Here are some related questions to deepen your understanding:

  1. How can the Intermediate Value Theorem be applied to functions with discontinuities?
  2. What are some common cases where the Intermediate Value Theorem does not apply?
  3. How can we estimate the value of the solution in each case using numerical methods?
  4. Can the Intermediate Value Theorem guarantee more than one solution in an interval?
  5. How does the continuity of a function affect the applicability of the IVT?

Tip: The IVT is useful not only for proving the existence of a solution but also for narrowing down intervals where a root might lie, which can then be solved more precisely using other methods like Newton's method or the bisection method.

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Math Problem Analysis

Mathematical Concepts

Calculus
Continuity
Intermediate Value Theorem

Formulas

f(x) = ln(x) - (4 - √x)
g(x) = e^x - (3 - 2x)

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 11-12, College