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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

(a) Prove that the equation has at least one real root. (b) Use your calculator to find an interval of length 0.01 that contains a root.

ln x = 3 – 2x

(a)

To determine

To prove: The equation lnx=32x has at least one real root.

Explanation

Theorem used: The Intermediate value Theorem

Suppose that if f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a)f(b). Then there exists a number c in (a, b) such that f(c)=N.

Proof:

Express the equation as, lnx3+2x=0.

Consider the function f(x)=lnx3+2x.

The logarithmic function defined on the interval (0,) and the polynomial function defined on Real numbers.

The function f(x) is a combination of polynomial and logarithmic function and it is defined on the interval (0,). So it is continuous everywhere on its domain (0,).

Without loss of generality, take the sub-interval [1,2].

In order to show that there is at least one root of the equation lnx3+2x=0 in the interval (1,2), it is enough to show that there is a number c between 0 and 1 for which f(c)=0.

Take a=1, b=2 and N=0

(b)

To determine

To find: An interval of length 0.01 that contains a root by using the calculator.

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