Calculus/Calculus - Rules of the Derivative
Examine the derivative of
f[x] = x - Sin[x]
to see whether the curve y = f[x] ever goes down.
Is it true that x >= Sin[x] for x >= 0 ?
Please help me how to solve this.
In the question, you need the following formula/information
1) derivative of x is 1
2) derivative of Sin[x] is Cos[x]
3) derivative of a sum/difference is the sum/difference of derivatives
4) if the derivative of f[x] is positive (or >=0) then the function is increasing (non-decreasing, i.e. doesn't ever go down)
5) Cos[x] is between the values -1 and 1 for any x.
With the above, f'[x] = ( x - Sin[x] )'
= (x)' - (Sin[x]) ' (using info 3)
= 1 - Cos[x] (using info 1 & 2)
But -1 < = Cos [x] =< 1 which means (after multiplying by -1)
-1 < = - Cos [x] =< 1 which implies (after adding a 1 everywhere) that
0 < = 1 - Cos[x] = < 2
So we have shown that f[x] is > = 0 (non-negative) hence by info (4), f[x} is non-decreasing.
So the curve y=f[x] never goes down.
To see if x > = Sin[x] , look at f, it is clearly equal to zero. Since f[x] never goes down,
f[x] remains above zero for x >=0. I.e. x - Sin[x] >=0 for non-negative values of x.
Now manipulate the inequality a little bit (i.e. add Sin[x] to both sides to get
x >= Sin[x] for x >=0.