Is the region in question , the tangent line and the X-axis or the Y-axis?
I am trying to set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the point ( -1, 1)
Now, i am aware this is an area problem so i probably have
would g(x) = x ^3 -2x ?
for the tangent line would it be in the form y = mx + b?
How would i find that? I have graphed the x ^3 -2x and made the point ( -1, 1) and can see the region for the area.
Would i need to take the dirivitive of x ^3 -2x and plug in -1 for x?
Hmmm, well it makes a difference in how you set up the integral and the limits of the integral. But going back to your general question, yes, you would take the derivative and then plug in -1 into the derivative. This value then is your slope of the tangent line and since you have the point (-1,1) you can then use y=mx+b to get the equation of the tangent line. Whether it is the X-axis or the Y-axis, f(x) would be the tangent line and g(x) would be if you decide to do the integral in terms of the variable x. One thing though, if it is the X-axis, you will have two integrals as the g(x) is the x-axis from to - and then from - to .
@HallsofIvy - I dont know!
if f '(-1) = 1. then, the point (-1, 1) and y = mx + b should be
would that be the f(x) for my other equation?
The graph would be a straight line up one and over one!
Yes, y = x + 1 is the equation of the tangent line.
Here is a graph of g(x) and the tangent line, done using WolframAlpha.
Added in Edit:
As Youkla points out below, the equation of the tangent line is y = x + 2 .