use the intermediate value theorem to show that there is a root of the equation in the specified interval

Question

e^^x=3-2x, (0,1)

Answer 1

From e^x = 3 − 2x, form the equation 3 − 2x − e^x = 0.

A Casio Graphing Calculator shows no breaks or discontinuities in y = 3 − 2x − e^x for 0 ≤ x ≤ 1. Also, y(0) = 2 and y(1) = -1.718281828. The Intermediate Value Theorem states that there is at least one value of x (call it "c") for 0 ≤ x ≤ 1 that will satisfy y(c) = 0 which does fall between y(0) = 2 and

y(1) = -1.718281828.

Computing more and more accurate values of x by Newton's Method For Approximation Of Equation Roots (starting with x = 0.5) and re-feeding each succeeding value of x into the formula

x − (3 − 2x − e^x)/(-2 − e^x) [x minus the ratio f(x)/f'(x) where f(x) = (3 − 2x − e^x) and f'(x) = (-2 − e^x)]

until the value of x stops changing will confirm that there is a highly accurate root of 3 − 2x − e^x = 0 at

x = 0.5942049585 which is in the domain of 0 ≤ x ≤ 1.

Answer 2

Let f(x) = e^x + 2x - 3

f(x) is continuous on the interval (0,1)

f(0) = -2 < 0

f(1) = e - 1 > 0

Since f is continuous on (0,1), and f(0) and f(1) have opposite signs, then according to the Intermediate Value Theorem, , f(c) = 0 for some number, c, between x = 0 and x = 1.

So, f(c) = e^c + 2c - 3 = 0

In other words, e^c = 3 - 2c