Now I think I have the correct answer, but am not 100% sure. Is it correct, and if not, where did I screw up? Cheers.
Question
Find a function f(x) such that the point (1, 2) is on the graph of y =f(x), the slope of the tangent line at (1, 2) is 3 and f’’(x)= x – 1.
My answer:
Equation of slope of the tangent at (1, 2) is y= 3x
f’’(x)= x – 1
thus:
f’(x)= ½x² – x + c
To find c, we need to solve:
f’(x)= ½x² – x + c = 3 at x = 1
c = 3½
Now we can find f(x):
f(x)= x³/6 – ½x²+ 3½x + c
At the tangent point (x= 1), y= f(x) has to touch the curve y = 3x:
c = 1/6
f(x)= x³/6 – ½x²+ 3½x + 1/6