Find f'' , intervals , and inflection ; please help the following question ?

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1 Answer
Mar 5, 2018

Please see below.

Explanation:

So, f(x) = 1/2x - sinx, is a pretty straightforward function to differentiate.

Recall that d/dx(sinx) = cosx, d/dx(cosx) = -sinx and d/dx(kx) = k, for some k in RR.

Hence, f'(x) = 1/2 - cosx.
Hence, f''(x) = sinx.

Recall that if a curve is 'concave up', f''(x) > 0, and if it is 'concave down', f''(x) <0. We can solve these equations fairly easily, using our knowledge of the graph of y =sinx, which is positive from an 'even' multiple of pi to an 'odd' multiple, and negative from an 'even' multiple to an 'odd' multiple.

Hence, f(x) is concave up for all x in (0,pi) uu (2pi, 3pi), and concave down for all x in (pi, 2pi).

Generally speaking a curve will have a point of inflection where f''(x) = 0 (not always - there must be a change in concavity), and solving this equation gives: x in {0, pi, 2pi, 3pi}.

We know from part b that there are changes in concavity at these points, hence (0,0), (pi, pi/2), (2pi, pi), and (3pi, 3pi/2) are all points of inflection.