Science:Math Exam Resources/Courses/MATH101/April 2010/Question 9
April 2010 Exam
|April 2010 Exam|
|For any real number x, define . Find the minimum value of g(x).|
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|We can consider x as a constant inside the integral|
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|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 = etdt so that du = dt and v = et. Hence
The last integral in the sum is:
The derivative is directly computed to be:
g'(x) = x(e2 − 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) = e2 − 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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