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For each function below, determine whether or not the Extreme Value Theorem applies. In each case,state what the Extreme Value Theorem tells us. (Be careful...)(a) f(x)-x sin (x) on [0, π](b) g(r)--2 -7x+4 on (-2,3)(c) h(x) = (x + 4)-, on [-1,5]

Question
For each function below, determine whether or not the Extreme Value Theorem applies. In each case,
state what the Extreme Value Theorem tells us. (Be careful...)
(a) f(x)-x sin (x) on [0, π]
(b) g(r)--2 -7x+4 on (-2,3)
(c) h(x) = (x + 4)-, on [-1,5]
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For each function below, determine whether or not the Extreme Value Theorem applies. In each case, state what the Extreme Value Theorem tells us. (Be careful...) (a) f(x)-x sin (x) on [0, π] (b) g(r)--2 -7x+4 on (-2,3) (c) h(x) = (x + 4)-, on [-1,5]

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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

 

 
 
 
 
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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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