The tangent line to the graph of the function f at x=2 is y=-x+3. find f(2) and f prime(2)?

I don't know what to do. I think fprime(2) is 1, but I'm not sure. I don't know how to start this at all :/

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- Sep 16th 2009, 01:38 PMVelvet LoveCalculus Derivative Help
The tangent line to the graph of the function f at x=2 is y=-x+3. find f(2) and f prime(2)?

I don't know what to do. I think fprime(2) is 1, but I'm not sure. I don't know how to start this at all :/ - Sep 16th 2009, 01:57 PMMatt Westwood
$\displaystyle f'(2)$ is actually -1, because that's the slope of the tangent to f. (It's a straight line, y=mx+c where m = -1 is the gradient, and c = 3 is where that line meets the line x=0, the y axis).

To find $\displaystyle f(2)$, note that as y=-x+3 touches the graph of f at the point where x=2, all you need to do is feed x=2 into the equation for the tangent at that point (which is given), and see what value of y you get out of the end. This is because at x=2, the graph of f(x) and its tangent are at the same point.