1. ## Derivatives in Calculus

I just recently got to a section with f and such. The professor hasn't taught us much of anything and I have just been having a lot of trouble.

Basically I have a graph which I attached, its a MS paint sketch but its still all the same. Basically what it says is "Organize these from biggest to smallest" using the graph. A small explanation would be nice.

0, f'(2), f(3)-f(2), and 1/2[f(4) - f(2)]

There is also a question asking for y=f(x) at (4,3) passes through (0,2), find f(4) and f(4). How exactly do you find these? I just wish there were some good book examples actually about this section...

2. Would it help if I told you that that is the graph of $\displaystyle f(x) = ln (x)$ ?

3. Yes but at the same time I dont know how to do the 2nd one

And I need some explanation as to why one is higher than the other rather than "i plugged this in"

4. Originally Posted by TastyBeverage
Yes but at the same time I dont know how to do the 2nd one
Can you clarify the 2nd question a bit? I'm not sure what the question is.

5. here is what my book says:

If the tangent line to y = f(x) at (4,3) passes through the point (0,2), find f(4) and f(4).

6. Originally Posted by TastyBeverage
here is what my book says:

If the tangent line to y = f(x) at (4,3) passes through the point (0,2), find f(4) and f(4).
Recall that the x coordinate is the input of a function, and the y coordinate is the output. We have been given a point on the curve (4,3), with an input and an output. One of those inputs just so happens to be 4. Therefore:

$\displaystyle f(4) = 3$

The derivative of a function is simply the slope of that function at any point. The slope at a certain point is the same as the slope of a line tangent to that point. So, if a line passes through both point (4,3) and point (0,2) and is tangent to the original point, then the derivative at that point (f'4), must be equal to the slope of that line. Recall that the slope of a line (m) is equal to:

$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

Therefore:

$\displaystyle f'(4) = \frac{2 - 3}{0 - 4} = \frac{1}{4}$