Find the point of inflection for f(x)=(x-2)/(x^2-3x-4)I need help finding the second derivative

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Indeed, to determine the inflection point of the function, we'll have to find out the 2nd derivative.

First, we need to find out the 1st derivative and for this reason, we'll use the quotient rule:

(u/v)' = (u'*v - u*v')/v^2

f'(x) = [(x-2)'*(x^2 - 3x - 4) - (x-2)*(x^2 - 3x - 4)']/(x^2 - 3x - 4)^2

f'(x) = [x^2 - 3x - 4 - (x-2)*(2x-3)]/(x^2 - 3x - 4)^2

f'(x) = (x^2 - 3x - 4 - 2x^2 + 7x - 6)/(x^2 - 3x - 4)^2

f'(x) = (-x^2 + 4x - 10)/(x^2 - 3x - 4)^2

We'll differentiate now f'(x):

f"(x) = {(-x^2 + 4x - 10)'*(x^2 - 3x - 4)^2 - (-x^2 + 4x - 10)*[(x^2 - 3x - 4)^2]'}/(x^2 - 3x - 4)^4

f"(x) =[(-2x + 4)*(x^2 - 3x - 4)^2 - 2(x^2 - 3x - 4)*(2x-3)*(-x^2 + 4x - 10)]/(x^2 - 3x - 4)^4

f"(x) = [(-2x + 4)*(x^2 - 3x - 4) - 2*(2x-3)*(-x^2 + 4x - 10)]/(x^2 - 3x - 4)^3

f"(x) = (-2x^3 + 6x^2 + 8x + 4x^2 - 12x - 16 + 4x^3 - 16x^2 + 40x - 6x^2 + 24x - 60)/(x^2 - 3x - 4)^3

f"(x) = (2x^3 - 12x^2 + 60x - 76)/(x^2 - 3x - 4)^3

f"(x) = 2(x^3 - 6x^2 + 30x - 38)/(x^2 - 3x - 4)^3

The second derivative of the function is f"(x) = 2(x^3 - 6x^2 + 30x - 38)/(x^2 - 3x - 4)^3.

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