Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. Ex = 6 − 5x, (0, 1) The equation ex = 6 − 5x is equivalent to the equation f(x) = ex − 6 + 5x = 0. F(x) is continuous on the interval [0

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Answer :

F(0) = -5  < 0

F(1) = e - 1 > 0

since the functions : f(0) and f(1) have opposite signs then there is a 'c' whereby F(c) = 0 ( intermediate value theorem fulfilled )

Hence there is a root in the given equation : [tex]e^x = 6 - 5x[/tex]

Explanation:

using Intermediate value Theorem

If F(x) is continuous and f(a) and f(b) have opposite signs then there will be a'c'E (a,b)  whereby F(c) = 0

given equation : [tex]e^x = 6 - 5x[/tex]   on (0,1)

and F(x) = [tex]e^x - 6 + 5x = 0[/tex]

This shows that the F(x) is continuous on (0,1)

F(0) = [tex]e^0 - 6 + 5(0)[/tex]  = -5 which is < 0

F(1) = [tex]e^1 -6 + 5(1)[/tex] = e -1 > 0  and e = 2.7182

since the functions : f(0) and f(1) have opposite signs then there is a 'c' whereby F(c) = 0 ( intermediate value theorem fulfilled )

Hence there is a root in the given equation : [tex]e^x = 6 - 5x[/tex]

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