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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[0], 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.


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