Calculus problem...please help?

Comments

• jeff-aaron

  May 2021

  y = (4x/1) + x^3 = 4x + x^3

  The point (1,2) is not on that curve.

  Or did you mean:

  y = (4x) / (1 + x^3)

  Let a = 4x, so a' = 4

  Let b = 1 + x^3, so b' = 3x^2

  So y = a/b, so we use the Quotient Rule:

  y' = (a'b - ab') / (b^2)

  y' = ((4)(1 + x^3) - (4x)(3x^2)) / ((1 + x^3)^2)

  y' = (4 + 4x^3 - 12x^3) / (x^6 + 2x^3 + 1)

  y' = (4 - 8x^3) / (x^6 + 2x^3 + 1)

  At x = 1, we have:

  y' = (4 - 8*1^3) / (1^6 + 2*1^3 + 1)

  y' = (4 - 8*1) / (1 + 2*1 + 1)

  y' = (4 - 8) / (1 + 2 + 1)

  y' = -4 / 4

  y' = -1

  So the slope of the tangent line is -1. So the tangent line is y = -x + b, which passes through (1,2), so we have:

  2 = -1 + b

  b = 2 + 1

  b = 3

  So the tangent line is y = -x + 3

  The y-intercept (where x = 0) is 3

  The x-intercept (where y = 0) is: -x + 3 = 0, so x = 3

  So if the tangent line is the hypotenuse of a right-angled triangle, and both axes are the legs, then the right angle is at the origin, so each leg has length 3. So the area of that triangle is (1/2) * 3 * 3 = 9/2 = 4.5