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March 28, 2017

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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 for a.

Yes, the point is (0,1/4) but it's not on the curve. It's on the tangent line. I'm not sure how to solve for a and a3 in that equation. My algebra is bad.
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How do you do this question.
f(x) = (5x-1)/(2-3x), find f^(-1)(x). The answer is supposed to be (1+2x)/(5+3x).
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And can you please explain this one to me.
Suppose that xy^(3)+2y^(2)+xy+2x^(3) = 0. Find the general expression for dy/dx. Also, find the value of dy/dx at the point where x = 1.
I have no idea how to differentiate that.

Thanks so much for the help!

  • Calculus - Damon - ,

    I get
    1) 3a^2=(a^3-1/4)/a

    3a^3=a^3-1/4
    2a^3= -1/4

    a^3=-1/8

    so a = -1/2

    Thus the line has gradient
    3(-1/2)^2 = 3/4

    and equation

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

    or

    4y=3x+1

  • Calculus - Damon - ,

    Thanks so much!

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