Please help me with Calculus?

Question

Let f be a function such that f"(x)=6x+8. 

 

Find f(x) if the graph of f is tangent to the line 3x-y=2 at the point (0, -2). 

 

 

Answer 1

Let's start by integrating f''(x) twice to get a general solution of f(x).  This yields:

 

f'(x) = 3x² + 8x + C

 

f(x) = x³ + 4x² + Cx + D

 

Now we solve for C and D using the information that it is tangent to the line 3x - y = 2 at the point (0, -2).

 

The fact that it is tangent means that f(x) must also hit the point (0, -2) and that the slope of f(x) at (0, -2) must be equal to the slope of the line 3x - y = 2 at the point (0, -2)

 

First, let's rearrange 3x - y = 2 --> y = 3x - 2

 

The slope of this line will always be 3

 

this means f'(0) = 3

 

Therefore C = 3

 

Finally, f(0) must equal -2

 

Therefore D = -2

 

The full equation: f(x) = x³ + 4x² + 3x - 2

Answer 2

If you take the integral of f"(x) you get:

f'(x) = 3x^2+8x+C

Taking a second integral will yield

f(x) = x^3+4x^2+Cx+D

We know that f(x) is tangent to y = 3x-2 at the point (0,-2), so we know that (0,-2) is a point on the graph:

f(0) = -2 = (0)^3 + 4(0)^2 + C(0) + D = D

Which implies that:

D = -2

We also know that at point x = 0, f'(x) is going to have the slope of y = 3x-2, which is m = 3.  Thus:

f'(0) = 3 = 3(0)^2+8(0)+C = C

Which implies that:

C = 3

So the final equation will be:

f(x) = x^3+4x^2+3x-2