Show that the equation sin(x)-3x+2=0 has exactly one real root.

Question

Show that the equation sin(x)-3x+2=0 has exactly one real root.

A) Let f(x)= sin(x)-3x+2

f(0)=____ (</>) 0

f(pi/2)=_____ (</>)0

B) fill in the blanks

Since sin x, 3x and 2 are continuous on R, then f (is/ is not) continuous on R. By the Intermediate Value Theorem there is a number c between 0 and pi/2 such that f(c)=0. Hence, sin (x)-3x+2=0 has a real root. To show that sin(x)-3x+2=0 has no other real roots, we will use Rolle's Theorem. Suppose that there are two real roots a and b. Then f(a)= f(b)=0. We've already shown that f is continuous on R, so f (is/ is not) continuous on [a,b]. f'(x)=______ is differentiable on R and so on [a,b]. Therefore, by Rolle's Theorem, there is a number k  (a, b) such that f'(k)=0. But f'(x)=0⇔_______=3, which is a contradiction! (-1≤ cos(x) ≤1). Therefore sin(x)-3x+2=0 has (many/ exactly one) solution.

Answer 1

sin x = 3x - 2

-1 ≤ sin x ≤ 1

-1 ≤ 3x - 2 ≤ 1

1 ≤ 3x ≤ 3

1 / 3 ≤ x ≤ 1

There are a unlimited number of solutions.