Okay Im REALLY confused with derivatives and how to compute them. THe problem I have is this:
There is a graph in which a straight line crosses -0.25 on the x-axis and 1 on the y-axis. Find f'(1/2).
What do I do?? Please help
Okay Im REALLY confused with derivatives and how to compute them. THe problem I have is this:
There is a graph in which a straight line crosses -0.25 on the x-axis and 1 on the y-axis. Find f'(1/2).
What do I do?? Please help
the derivative gives you the formula for the slope. once you find the derivative and plug in the required x-value and that gives you your m in the equation of a line (y = mx + b), you can then use your m and a point the line passes through to get your tangent line. however, that is not the case here. you are required to find a slope, why would you find the tangent line. that's not what they asked for, and you don't need it to find the slope. (the tangent line for a straight line, at any point on the line, is itself, by the way). all you are doing is finding the slope of a line that passes through (-0.25, 0) and (0, 1). the slope (for all points on the straight line) is just given by $\displaystyle m = \frac {y_2 - y_1}{x_2 - x_1}$, and you know what those components are, right?