SP1 (post: 1454530) wrote:I imagine he is trying to get you to notice that the slope of the secant line approaches the slope of the tangent as the two points chosen become closer and closer to the point P. Since the derivative (slope) of this curve is -1/(x*x), then the slope at P is -1/4 or -0.25. Notice that the points in the previous part show a slope closer to -0.25 when the two points are nearer to P. By algebra, I assume he means using the 2-point solution to slope where:

slope = (y2-y1)/(x2-x1)

Hope this helps.

Ok, thanks. I think I understand that one now, sorta.

Now I'm stuck on a problem involving continuity that I think should be fairly easy, but I'm not sure.

If f and g are continuous functions with f(2)=4 and, find g(2).

Shouldn't the answer just be 4? If they are continuous with it, they would be the same, correct? But plugging in 4 for both values doesn't work out with that equasion.

Edit: Would it be .875? If f(x) and g(x) both multiply by 2 (which I guessed that they do since f(2) = 4 and they are continuous with f(2)=4), .875*2 = 1.75, and when you plug that into the equasion you get 7...but for some reason I think I did something wrong there. The other answer I came up with was g(2) = -5. Which is completely different.

Edit2: Also, I have no clue where to even start with these. I don't really want an answer, because thusfar I haven't even been able to make an attempt at them, but I don't even know how to approach these.

[quote]

3. Find all constants c that make f continuous on (-∞]

Edit3: Ok, I think I got number 3. The answers are 1 and -5. But I figured those out through trial and error, is there a better way to do that?