Find the absolute extrema if they exist as well as where they occur. f(x)= x-2 /

Question

Find the absolute extrema if they exist as well as where they occur. f(x)= x-2 / x^2+3x+6

Explanation / Answer

Tendency at both ends of the curve Because you are asking about absolute extrema, you need to know where the function is heading when x approaches infinities. e.g. If the function value approaches inifinity also, you dont get any absolute extrema. Since the order of numerator is lower than denominator, f(x) -> 0 when x -> oo and x-> -oo (oo = infinity) (2) locate points with zero slope First find the first derivative. (I'm used to using the product rule + chain rule instead of quotient rule when working out derivatives of fractions. I do it by rewriting the denominator as (xxxxx) ^ -1 ) f'(x) = (x-2) (-1) (x^2 + 3x + 6)^-2 (2x+3) + (x^2 + 3x + 6)^-1 (1) = [ -(x-2) (2x+3) + (x^2 + 3x + 6) ] (x^2 + 3x + 6 )^-2 = (-x^2 + 4x + 12) / (x^2 + 3x + 6 )^2 when f'(x) = 0, -x^2 + 4x + 12 = 0 (x+2)(x-6) = 0 x = -2 or x = 6 (with one condition: the denominator x^2 + 3x + 6 not zero. but the delta of this = b^2 - 4ac = 9 - 4*1*6 < 0, so the denominator cannot be zero.) Using First derivative test to see if they are maxima / minima / pt of inflexion. When x = -2, it is minimum. y = -1 When x = 6, it is maximum. y = 1/15

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