Given that the second derivative of the curve is 6x, what is the equation of the curve? The line y=x+2 is tangent to the curve at point (1,3).

Answer 1

Taking the antiderivative of the second derivative will give you the first derivative.

Antiderivative of 6x is 3x² + C

First derivative is 3x²+C. The first derivative of a point is the slope of the tangent line at that point.

The slope of the tangent line y=x+2 is 1 ( compare it with y=mx+b)

Now that we have the slope and the point (1,3), we can find the value of the constant in the first derivative.

3x²+C= slope

3(1)²+C=1

3+C=1

C=-2

First derivative = 3x²-2

Now we can find the anti derivative of the first derivative to find the original function

The anti derivative of 3x²-2 = x³-2x+C

f(x)= x³-2x+C

To find the value of C we can use the point (1,3)

f(1)= (1)³-2(1)+C

3=1-2+C

C=4

So the original function f(x)= x³-2x+4

Answer 2

integral of 6x = 3x^2 + C

integral of 3x^2 + C = x^3 + Cx + K

the 1st derivative = slope of the tangent line = 1 = 3x^2 + C = 1=3(1)+ C, C =-2

x^3 -2x + K = 3 when x = 1

1-2+K = 3

K= 3+1 = 4

the equation of the curve is x^3 -2x + 4