Let a function f:[0,5]R be continuous, f(1)=3 and F be defined as: F(x)=_1^xt^2g(t),dt, wh…

Question

Let a function $f:[0,5]	o\mathbb{R}$ be continuous, $f(1)=3$ and $F$ be defined as: $F(x)=\displaystyle\int_1^xt^2g(t)\,dt$, where $g(t)=\displaystyle\int_1^tf(u)\,du$. Then for the function $F$, the point $x=1$ is:

Accepted Answer

By Leibnitz's rule, $F'(x)=x^2g(x)=x^2\displaystyle\int_1^xf(u)\,du\ \ldots(i)$

At $x=1$: $F'(1)=1\cdot\displaystyle\int_1^1f(u)du=0$.

Differentiating (i): $F''(x)=x^2f(x)-2x\displaystyle\int_1^xf(u)\,du$... more precisely $F''(x)=x^2f(x)+2xg(x)$

At $x=1$: $F''(1)=1\cdot f(1)+2\cdot1\cdot g(1)=f(1)+0=3$

Since $F'(1)=0$ and $F''(1)=3>0$, $x=1$ is a point of local minima.

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