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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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