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

Question

Let a function f : [0, 5] $$ \to $$ R be continuous,
f(1) = 3 and F be defined as :

$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$

Then for the function F, the point x = 1 is :

Accepted Answer

$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$

$$ \Rightarrow $$ F'(x) = x2g(x) = x2$$\int\limits_1^t {f(u)du} $$

$$ \therefore $$ F'(1) = (1)(0) = 0

Now, F''(x) = 2xg(x) + x2g'(x)

F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3

[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]

So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0

$$ \therefore $$ For the function f(x), x = 1 is a point of local minima.

Solutions · 1