Given, F(x)=∫1xt2g(t)dt
By Leibnitz rule we get,
F′(x)=x2g(x)
⇒F′(1)=1.g(1)=0 (∵g(1)=0)
Now F′′(x)=2xg(x)+x2g′(x)
⇒F′′(x)=2xg(x)+x2f(x) (∵g′(x)=f(x))
⇒F′′(1)=0+1×3
⇒F′′(1)=3
F(x) has a local minimum at x=1.
Let a function f:[0,5]→R be continuous, f(1)=3 and F be defined as:
F(x)=∫1xt2g(t)dt, where g(t)=∫1tf(u)du.
Then for the function F(x), the point x=1 is:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
a point of local minima
not a critical point
a point of local maxima
a point of inflection
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