If the graph of f passes through the point
(1, 0) and the slope of the tangent line at(x, f(x)) is 4x − 3, find the value of f(2).
To find the value of f(2), we need to use the information given about the graph of f and the slope of the tangent line.
We know that the graph of f passes through the point (1, 0). This means that when x = 1, f(x) = 0. So we have f(1) = 0.
Next, we are given that the slope of the tangent line at (x, f(x)) is 4x - 3. This means that the derivative of f(x) with respect to x is equal to 4x - 3.
To find f(2), we need to integrate the derivative of f(x) to get the original function f(x). Let's integrate 4x - 3:
∫(4x - 3) dx = 2x^2 - 3x + C
Now we need to find the value of C, which represents the constant term. To do this, we can use the fact that f(1) = 0. Let's substitute x = 1 and f(x) = 0 into the equation:
2(1)^2 - 3(1) + C = 0
2 - 3 + C = 0
-1 + C = 0
C = 1
So now we have the function f(x) = 2x^2 - 3x + 1.
To find f(2), we substitute x = 2 into the equation:
f(2) = 2(2)^2 - 3(2) + 1
f(2) = 2(4) - 6 + 1
f(2) = 8 - 6 + 1
f(2) = 3
Therefore, f(2) = 3.