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anonymous

  • 5 years ago

A and B are the points on the curve y=2^x with the coordinates 3 and 3.1 respectively. Find the gradient/slope of the chord AB and stating the points you use, find the gradient of another chord which will give a closer approx. to the gradient/slope of the tangent to y=2^x at A. thanks

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  1. amistre64
    • 5 years ago
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    A = (3, 2^3) = (3,8) B = (3.1,2^(3.1)) = (3.1, 8.57) slope = .57/.1 = 5.7

  2. amistre64
    • 5 years ago
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    since a is at 3; use a closer value of x that is between 3 and 3.1; such as 3.01

  3. amistre64
    • 5 years ago
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    as you get closer and closer to A, which is (3,8); you will approach the limit that is equal to the slope at 'A'; also known as the derivative of y = 2^x at x=3

  4. amistre64
    • 5 years ago
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    does that help?

  5. anonymous
    • 5 years ago
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    wait let me read it through a few times :S

  6. anonymous
    • 5 years ago
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    how did you get the slope? where did you get the numbers from 0.57/0.1=5.7???

  7. amistre64
    • 5 years ago
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    you are given a function for y: 2^x correct? At point A it says that x=3 and at point B, x=3.1 right?

  8. anonymous
    • 5 years ago
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    yes

  9. amistre64
    • 5 years ago
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    point A = (3,2^3) point B = (3.1, 2^(3.1))

  10. amistre64
    • 5 years ago
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    B-A gives you the slope: x , y -------------- 3.1 , 2^(3.1) -3 , - 2^3 --------------- ( .1 , .57418...) slope = y/x = .57/.1 = 5.7/1 = 5.7

  11. anonymous
    • 5 years ago
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    thanks

  12. amistre64
    • 5 years ago
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    youre welcome :)

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