Few AP Questions

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  • #1
razored
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Find the slope of the line containing the point (4,0) and tangent to the graph of y=e^(-x)


: I found the derivative (-e^(-x)) and substituted x=4. I got an answer of -0.018 but the answer is -0.050.. what did i do wrong?


If http://texify.com/img/%5CLARGE%5C%21%5Cint_%7B0%7D%5E%7B3%7D%20%28x%5E2%20-4x%20%2B4%29dx%20.gif [Broken] is approximated by 3 inscribed rectangles of equal width on the x-axis, then the approximation is :


I got the answer of 5 but it is apparently 1. I did : 1(4)+ 1(1)+(1)(0) where the width of the rectnagles is 1 and and the second values are the respective y values at x.

What mistake did I make here as well?
 
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Answers and Replies

  • #2
HallsofIvy
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The tangent line passes through (4,0). It is not tangent to the curve when x= 4. Assume the line is tangent to the curve at, say, x= a and write the equation for the tangent line in terms of a. Set x= 4 and y= 0 and solve for a.

As far as your second question is concerned, I don't see how we can say what you did wrong when you don't say what you did! Of course, the result depends upon exactly where you evaluate the function in each interval so there are an infinite number of correct answers. What do you get if you choose the right endpoint of each interval to evaluate the function?
 
  • #3
razored
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I got the answer of 5 but it is apparently 1. I did : 1(4)+ 1(1)+(1)(0) where the width of the rectnagles is 1 and and the second values are the respective y values at x.
I was told to use 3 rectangles of equal width on the x-axis implying delta x =1 for each rectangle. Then I used the left-end point method. Consequently, (1)f(0) + 1f(1) + 1f(2) = 1(4)+ 1(1)+(1)(0) = 5 but the answer is 1. I don't see why.


For the first one, i did http://texify.com/img/%5CLARGE%5C%21%5Cfrac%7Be%5E%7B-x%7D%7D%7Bx-4%7D%3D-e%5E%7B-x%7D.gif [Broken]which[/URL] I found x to be 5. I substituted 5 into -e^(-x) and that does not equal -.05 . What did I do wrong now?
 
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