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Problem of the Day No calculator!

Problem of the Day No calculator!. What is the instantaneous rate of 
change at x = 2 of f(x) = x2 - 2 ? x - 1. A) -2 C) 1/2 E) 6 B) 1/6 D) 2. Problem of the Day No calculator!.

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Problem of the Day No calculator!

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  1. Problem of the Day No calculator! What is the instantaneous rate of 
change at x = 2 of f(x) = x2 - 2 ? x - 1 A) -2 C) 1/2 E) 6 B) 1/6 D) 2

  2. Problem of the Day No calculator! What is the instantaneous rate of 
change at x = 2 of f(x) = x2 - 2 ? x - 1 A) -2 C) 1/2 E) 6 B) 1/6 D) 2 (take derivative and then substitute in)

  3. Newton's Method A technique for approximating the real 
zeroes of a function using tangent lines

  4. Newton's Method A technique for approximating the real 
zeroes of a function using tangent lines y x a b If the function is continuous on [a, b] and 
differentiable on (a, b) and if f(a) and f(b) differ in sign then by the ___________________________ 
f must have at least one zero in (a, b)

  5. Newton's Method A technique for approximating the real 
zeroes of a function using tangent lines y x a b If the function is continuous on [a, b] and 
differentiable on (a, b) and if f(a) and f(b) differ in sign then by the Intermediate Value Theoremf must have at least one zero in (a, b)

  6. Newton's Method A technique for approximating the real 
zeroes of a function using tangent lines Visual Calculus Link

  7. Newton's Method A technique for approximating the real 
zeroes of a function using tangent lines In summary, the x-intercept will be approximately xn+1 = xn - f(xn) f '(xn)

  8. Calculate 3 iterations of Newton's 
Method to approximate a zero of f(x) = x2 - 2 starting with x = 1. f(xn) f '(xn) f(xn) f '(xn) xn - xn f '(xn) f(xn) Iteration

  9. Calculate 3 iterations of Newton's 
Method to approximate a zero of f(x) = x2 - 2 starting with x = 1. f(xn) f '(xn) f(xn) f '(xn) xn - f '(xn) xn f(xn) Iteration -.5 .083 .002451 2 3 2.83 1.5 1.416 1.414216 1 2 3 1 1.5 1.416 -1 .25 .006945

  10. Calculate 3 iterations of Newton's 
Method to approximate a zero of f(x) = x2 - 2 starting with x = 1. f(xn) f '(xn) xn - Ti-84 xn Iteration Y1 = your equation Y2 = nderiv(Y1,x,x) x - Y3 = Ti-Nspire f1 = your equation f2 = f3 =

  11. Newton's Method will not always 
produce an answer, such as when 1) the derivative within the interval is zero at any point 2) functions similar to f(x) = x1/3

  12. You can test for convergence to see if it 
will work with the following formula f(x) f ''(x) [f '(x)]2 < 1

  13. Another precaution Do not round in intermediary steps. Let your calculator carry the numbers.

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