Atachement
Recall that if we have y = f(g(x)) that
And if y = f(g(h(x)))
etc, etc.
So to facilitate this process a bit I'm going to rewrite this as:
We are going to be using the product rule: y = f(x)g(x) --> y' = f'(x)g(x) + f(x)g'(x). Then we use the chain rule on the individual factors.
After a bit of simplifying I get:
Now. At the point (2, 4) the slope has a value of -93/2. Thus the tangent line will be:
and will pass through the point (2, 4):
Thus the tangent line is
-Dan
Like the first problem we need to have the value of the slope of the function. So:
I'll use the quotient rule here, then the chain rule for each factor:
After a bit of simplification:
Now for the first tangent line. We want the tangent at the point . The slope will be:
So the first line has the form . Inserting the point into this I get that .
So the first line is: .
In a similar fashion the line tangent to the function at the point is: .
To find the point of intersection we need to solve the system of equations:
Use your favorite method here. It's ugly and if you want help I'll provide it. Until then I'll simply state the solution:
The point of intersection is: .
Two comments:
1) Check my math. This is sufficiently complicated I may have made a small error somewhere. I'd simply graph the problem, but my nicer graphing software isn't available. (On my TI-92 it seems to check out, but the resolution is bad.)
2) Whoever came up with this one is a sadist!
-Dan
I'm presuming this message is directed at me. I'm going to post the solution anyway, just in case someone wants it.
Note: A graphing calculator would be very handy here. If you don't have one I would recommend using decimals instead of trying to combine the fractions by hand. (This is one of the rare places I suggest that.)
We are looking for a place where x and y are the same for both equations. Since both are solved for y, we can say that the right hand sides of both equations are equal:
Then use either of the original equations to find y. I'll use the first:
-Dan