Thread: Is this right? Finding function f(x) given tangent slope and f''(x)

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

Originally Posted by Dr Zoidburg

Now we can find f(x):
f(x)= x³/6 – ½x²+ 3½x + c

At this point use the fact that (1,2) is on the graph. That is f(1)=2 to get
f(1)= 1/6 – ½1²+ 3½1 + c = 2. This gives me

D'oh!
I made f(x) equal 3 (the slope) and not 2 (the y intercept).
Many thanks for spotting that mistake for me. Much appreciated.