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

A) Let f(x) = sin(x) – 3x + 2

f(0)=  2  > 0

f(π/2) =  1 – (3π/2) + 2  ≈ –1.71 < 0

B) fill in the blanks

Since sin x, 3x and 2 are continuous on R, then f  is  continuous on R. By the Intermediate Value Theorem there is a number c between 0 and π/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  continuous on [a,b]. f'(x) =   cos(x) – 3   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 ⇔   cos(x)   = 3, which is a contradiction! (–1 ≤ cos(x) ≤ 1). Therefore sin(x) – 3x + 2 = 0 has  exactly one  solution.