Question

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. ex = 3 − 2x, (0, 1) The equation ex = 3 − 2x is equivalent to the equation f(x) = ex − 3 + 2x = 0. f(x) is continuous on the interval [0, 1], f(0) = _____, and f(1) = _____. Since f(0) < 0 < f(1) , there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation ex = 3 − 2x, in the interval (0, 1)

Answer 1

Answer:

f(x) is continuous on the interval [0, 1], f(0) = -2 , and f(1) = 1.718 . Since f(0) < 0 < f(1) , there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation ex = 3 − 2x, in the interval (0, 1)

Step-by-step explanation:

From the question we are told that

   The equation is $f(x) = e^x - 3-2x$  

    The interval is  [0, 1]

Generally f(0) is  

      $f(0) = e^0 - 3-2(0)$

=>   $f(0) = 1 - 3-2(0)$  

=>    $f(0) = -2$  

Generally f(1) is

       $f(1) = e^1 - 3-2(1)$

       $f(1) = e^1 - 1$

       $f(1) = 1.718$

From the value we see that at x =  0  , f(0) =  -2 which is below the x-axis

    and  the  at  x = 1  , f(1) =  1.718  which is above the x-axis

Now  the according to Intermediate Value Theorem , given the condition stated above, there will exist a root c in the interval  such that

    f(c) =  0