Use the Intermediate Value Theorem to show that there is a solution of the given equation in the specified interval.

Question

Use the Intermediate Value Theorem to show that there is a solution of the given equation in the specified interval.

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

Answer 1

Consider e^x - 3 + 2x = 0

When x = 0, e^x - 3 + 2x = 1 - 3 + 0 = -2 < 0

When x = 1, e^x - 3 + 2x = e¹ - 3 + 2 = e¹ - 1 > 0

e^x - 3 + 2x is a continuous function

So as it is <0 when x = 0, and > 0 when x = 1, by the IVT there must be at least one value of x in the interval (0,1) when it is 0 . When it is zero, e^x = 3 - 2x

Mike

Answer 2

We are trying to show that there is a value in (0, 1) such that e^x = 3 - 2x, or e^x - (3 - 2x) = 0, or e^x + 2x - 3 = 0

At 0, e^x + 2x - 3 = 1 + 0 - 3 = -2< 0

At 1, e^x + 2x - 3 > 2.718 + 2 - 3 = 1.718 > 0

As f(0) < 0 and f(1) > 0 and f(x) is continuous, there is a value in (0, 1) such that f(x) = 0, or

e^x = 3 - 2x