calculus

posted by .

Find a function f such that the curve y = f(x) satisfies y'' = 12x, passes through the point (0,1), and has a horizontal tangent there.

I can use the anti-derivatives to get

y = 2x^3 + Cx + D

but I don't know how to get C even though I know D = 1.

  • calculus -

    you know that
    y' = 6x + C
    and that y'(0) = 0, so
    0 = C

    y = 2x^3 + 1

  • calculus -

    So y'(0) = 0 because there is a horizontal tangent at x = 0?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus - Damon

    Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point. So I found the derivative which is 3x^2. Let (a, a3) be the point of tangency. 3x^2 = (a3 - 1/4)/(a-0) I'm not sure how to solve …
  2. Help Calc.!

    original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point …
  3. please help; calc

    original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point …
  4. Need help fast on calc

    original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point …
  5. calculus

    Suppose y is defined implicitly as a function of x by x^2+Axy^2+By^3=1 where A and B are constants to be determined. Given that this curve passes through the point (3,2) and that its tangent at this point has slope -1, find A and B.
  6. calculus

    Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each horizontal tangent line to the curve. c. The line through the origin with slope -1 is tangent to the curve …
  7. calculus

    A curve passes through the point (1,-11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,-16) is parallel to the x-axis. Find i) the values of a and b ii) the equation …
  8. calculus (steps and explanation plz)

    Find a curve y=f(x) with the following properties (i) y"=6x and (ii) its graph passes through the point (1,1) and has a horizontal tangent there.
  9. calculus

    find the equation of the curve for which y''=(12/x^3) if it passes through (1,0) and is tangent to the line 6x+y=6 at that point.
  10. Calculus AB

    Could someone please help me with these tangent line problems?

More Similar Questions