Given that the graph of f passes through the point (1, 14) and that the slope of its tangent line at (x,f(x)) is 4x+5 find f(3).
2007-05-01 11:06:21 · 3 answers · asked by Pretty Girl 1 in Science & Mathematics ➔ Mathematics
So.... f '(x) = 4x + 5
Then, find the antiderivative...
f(x) = 2x^2 + 5x + C
Plug in (1, 14) to find C.
14 = 2(1)^2 + 5(1) + c
14 = 7 + C
7 = C
So... f(x) = 2x^2 + 5x + 7
So, then...
f(3) = 2(3)^2 + 5(3) + 7 = 18 + 15 + 7 = 40
2007-05-01 11:11:27 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
Since the slope is 4x+5, f(x) is 2x^2+5x+c=y. But since (1,14) is on this graph, 14=2*(1)^2+5(1)+c--or c=7. So f(3)=40.
2007-05-01 18:11:35 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋
f'(x) = 4x+5
f(x) = 2x^2 + 5x + C
Solve for C using the data point supplied (f(1) = 14):
f(1) = 14
2(1)^2 + 5(1) + C = 14
7 + C = 14
C = 7
f(x) = 2x^2 + 5x + 7
2007-05-01 18:11:58 · answer #3 · answered by McFate 7 · 0⤊ 0⤋