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Question 09 

For any real number x, define . Find the minimum value of g(x). 
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Hint 

We can consider x as a constant inside the integral 
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Solution 

First we solve the integral considering x as a constant:
The first integral in the sum is
Now we solve by parts letting u = t and dv = e^{t}dt so that du = dt and v = e^{t}. Hence
The last integral in the sum is:
So
The derivative is directly computed to be: g'(x) = x(e^{2} − 1) − 2 Which is zero at . To make sure it is a minimum we use the second derivative test. The second derivative is: g''(x) = e^{2} − 1 Which is always positive (Remember that e > 2). This means is the minimum for g. Plugging in this value gives
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