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.
--------------------------------------
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).
--------------------------------------
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!

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

Thanks so much!

For the first question:

To find the equation of the tangent line that passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point, you're on the right track by finding the derivative of the curve.

The derivative of y=x^3 with respect to x is given by dy/dx = 3x^2.

Let (a, a^3) be the point of tangency. We want to find the value of a.

Since the tangent line passes through the point (0, 1/4), we can substitute these values into the equation of the tangent line to get:

3x^2 = (a^3 - 1/4)/(a - 0)

Simplifying this equation, we get:

3x^2 = (a^3 - 1/4)/a

To solve for a, we can cross-multiply:

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

Rearranging the equation, we have:

a^3 - 3x^2 * a - 1/4 = 0

Now, you can try to factorize or use numerical methods to solve for a.

For the second question:

To find f^(-1)(x), we need to find the inverse function of f(x) = (5x-1)/(2-3x).

To do this, we start by replacing f(x) with y:

y = (5x-1)/(2-3x)

Next, swap x and y:

x = (5y-1)/(2-3y)

Now, solve this equation for y.

Cross-multiply to eliminate the fractions:

x(2-3y) = 5y-1

Expand:

2x - 3xy = 5y - 1

Group the y terms:

2x + 1 = 5y + 3xy

Rearrange:

3xy + 5y = 2x + 1

Factor out y:

y(3x + 5) = 2x + 1

Finally, solve for y:

y = (2x + 1)/(3x + 5)

Therefore, f^(-1)(x) = (1+2x)/(5+3x).

For the third question:

To differentiate the equation xy^3 + 2y^2 + xy + 2x^3 = 0 with respect to x, we need to apply the rules of differentiation.

Differentiating each term of the equation, we have:

d(xy^3) + d(2y^2) + d(xy) + d(2x^3) = d(0)

Using the product rule of differentiation, we get:

y^3 + xy^2(dy/dx) + 4y(dy/dx) + y + x(dy/dx) + 6x^2 = 0

Now, we can solve for dy/dx:

dy/dx = -(y^3 + y)/(xy^2 + 4y + x + 6x^2)

To find the value of dy/dx at the point where x = 1, substitute x = 1 into the equation above:

dy/dx = -(y^3 + y)/(y^2 + 4y + 1 + 6)

Simplifying further:

dy/dx = -(y^3 + y)/(y^2 + 4y + 7)

So, to find the value of dy/dx at the point where x = 1, we need to know the corresponding y-value or solve for y using additional information or methods.