Find the minimum value of the function, f(x)= x^2 + 5x + 2, where x belongs to the set of Real numbers
We first differentiate f(x), and we get f'(x)=2x + 5. We then set this equal to 0 and then solve for x. We get that xmin= -2.5. We check whether this was indeed a minimum, by calculating the second derivative, f''(xmin)= 2. Since f''(x) > 0 we know that xmin is indeed a (local) minimum. Then to find the minimum value of f(x), we substitute the value of x back to the equation and get the minimum value of f(x) is -4.25 ((-2.5)^2 + 5(-2.5) + 2 = -4.25))
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