Maxima and MinimaHard

Question

f(x) = and g(x) = f(t) dt, x ∈ [1, 3] then g (x) has

Options

A.local maxima at x = 1 + ln 2 and local minima at x = e
B.local maxima at x = 1 and local minima at x = 2
C.no local maxima
D.no local minima

Solution

g′(x) = f(x) =
g′(x) = 0, when x = 1 + ln2 and x = e
g′′(x) =
g′′(1 + ln 2) = - eln2 < 0 hence at x = 1 + ln 2, g(x) has a local maximum
g′′(e) = 1 > 0 hence at x = e, g(x) has local minimum.
∵   f(x) is discontinuous at x = 1, then we get local maxima at x = 1 and local minima at x = 2.

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