Question

Step 1

Given function f(x) = xsinx on [0 π]

According to Extreme value theorem if f(x) is continuous in [0 π] then it has one maximum and minimum value on [0 π].

In the given interval we know that f(x) is continuous from [0 π].

If we differentiate f(x) we get f'(x) = xcosx+sinx which is 0 at x = 0, where we get local minimum and at x = 84.87 radians f'(x) = 0 and this is a point for local maximum.

So we can say that for f(x) is continuous in [0 π] then there exist a local maximum and local minimum with iin this interval which obeys the extreme value theorem

Step 2

Given g(x) = x3 -2x2 -7x+4 on (-2,3).

According to extreme value theorem if g(x) is continuous on (-2,3) then there exist a local maximum and local minimum in (-2,3).

We can see that the function is continuous in (-2,3) as it is defined at eeach and every point.

g(x)=x3 -2x2 -7x+4 . So differentiating it with respect to x we get g'(x) = 3x2-4x-7.

g'(x) = 3x2-4x-7 = (x-7/3)(x+1) . We know that g'(x) is zero at 7/3 and at -1.

g''(x)=6x-4 . g''(x) is ...

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