Question

Prove that f(x)=2x-1-sin(x) has exactly one real root.

Answer 1

I understand how to show that there's a root in a given interval [a,b] where both a and b are x values with the same y value, but how do I prove that there exists only ONE root?

Answer 2

Let f(x) = 2x - 1 - sin(x)
Also, notice that, f(0) <0 and f(2)>0 Therefore, by the IVT there exists a zero in the interval [0, 2].

Answer 3

Now to find whether it's just one root, take the derivative and see if it's >0 for all x to see if your function is increasing or decreasing.

Answer 4

Wait, via Rolle's Therom, wouldn't there have to exist a c in [a,b] such that f'(c)=0?

Answer 5

If it had multiple roots.

Answer 6

ye only if it has multiple roots; but not necessarily in the interval [a, b]

Answer 7

somewhere, the function must "return back" to zero, causing the slope to be zero at some point, not necessarily close to any root

Answer 8

Okay, I think I got it. so \[-\cos(x)\neq2 \therefore\]
There's only one real root?

Answer 9

Yes.

Answer 10

Thanks @abb0t and @Euler271

Answer 11

Refer to the Mathematica calculation attached.