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.
you know that
y' = 6x + C
and that y'(0) = 0, so
0 = C
y = 2x^3 + 1
So y'(0) = 0 because there is a horizontal tangent at x = 0?
To find the constant C, you need to use the information given in the problem. We know that the curve has a horizontal tangent at the point (0,1).
A horizontal tangent means that the slope of the curve at that point is zero. This can be represented by setting the derivative of the curve with respect to x, y', equal to zero.
Taking the derivative of y = 2x^3 + Cx + D, we get:
y' = 6x^2 + C
Setting y' = 0, we have:
0 = 6x^2 + C
Now, substitute the x-coordinate of the given point (0,1) into this equation:
0 = 6(0)^2 + C
0 = C
Therefore, C = 0.
Now we can rewrite the equation for the curve as:
y = 2x^3 + Dx + 1
So the function f(x) that satisfies y'' = 12x, passes through the point (0,1), and has a horizontal tangent at that point is:
f(x) = 2x^3 + Dx + 1.