at which value of x does y=(x+1)(x-19)+7 takeon its minimum value?

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  1. y=f(x)= x^2-19x+x-19+7= x^2-18x-12
    Find the critical value:
    f'(x)= 2x-18, set it equal to zero
    0= 2x-18
    hence, x=9 is the critical value
    Find the second derivative:
    f"(x)= 2 & f"(9)= 2, since the second derivative is positive, the function has a minimum at x=2.
    x=2 is your answer.

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