Given that the graph of f ( x ) passes through the point ( 8 , 10 ) and that the slope of its tangent line at ( x , f ( x ) ) is 4 x + 2 , what is f ( 3 ) ?

Question

Given that the graph of  passes through the point  and that the slope of its tangent line at  is , what is ?

Answer 1

Since the slope is 4x + 2, then f'(x) = 4x + 2. Integrating both sides (taking the anti-derivative), we get f(x) = 2x² + 2x + C. Since the graph goes through (8, 10) then f(8) = 10 or:

10 = 2(8)² + 2(8) + C

10 = 128 + 16 + C

C = -134 so:

f(x) = 2x² + 2x - 134 and

f(3) = 2(3)² + 2(3) - 134 = -110

Answer 2

Hope that this file transfers ok. Peter. I wrote it out for you but the file wouldn't transfer. The solution is -110 and it is obtained by indefinitely integrating f'(x) and getting f(x) including a constant of C and then plugging in (8,10), solving for C which is -134, then evaluating f(x) = 2x^2 + 2x - 134 at x = 3. Then just to be sure check that all conditions are met. They are.