Show that the equation below has exactly one root Show all y

Show that the equation below has exactly one root. Show all your work. (Hint: Use Intermediate Value Theorem to show existence of at least one root and Apply Rolle\'s Theorem to show that there cannot be exactly 2 roots) 3x + 2 cos(x) + 5 = 0

Solution

The intermediate value theorem states that given a function f(x)

continuous on [a, b] and then there exists c in(a, b) such that:

f(a) < f(c) < f(b) or f(b) < f(c) < f(a).

Let:

f(x) = 3x + 2cos(x) + 5

This is a sum of continuous function, so the function itself is continuous.

f(-3) = -9 + 2cos(-3) + 5 < 0

f(0) = 2 + 5 = 7 > 0

Using the intermediate value theorem, I know there exists c in (-3, 0)

sucht that f(c) = 0, i.e. f(-3) < f(c) < f(0)   

This shows that it has at least one root.

Now, we need to show that it has only that root.

Rolle\'s theorem states that given a function

continuous on [a, b] and differentiable in (a, b)

and f(a) = f(b)

then there exists c in (a, b) such that:

f\'(c) = 0.

f(x) = 3x + 2cos(x) + 5

Assume that this has two roots x = a and x = b.

i.e. f(a) = f(b) = 0

f\'(x) = 3 - 2sin(x)

f\'(c) = 3 - 2sin(c)

Well, sin(c) is bounded by 1, so this means that:

-2sin(c) is between -2 and 2.

This thus mean that:

f\'(c) > 0 for all values of c.

Hence there exists no such c.

Hence there can\'t be more than one root.

This thus means:

f(x) = 3x + 2cos(x) + 5 has only one real root.