Question

Use Newton’s Method to find a solution of the equation e 6x+ 3(ln 2)2 e 2x− e 4x ln 8 − (ln 2)3 = 0 with error tolerance 10−5 , and that is in the interval −1 ≤ x ≤ 0.

Answer 1

Answer:

  x ≈ -0.746756

Step-by-step explanation:

We have interpreted your equation to be ...

  $e^{6x}+3\ln{(2)}^2e^{2x}-e^{4x\ln{(8)}}-\ln{(2)}^3=0$

A graphing calculator shows the desired zero to be near -0.747. Using that value as the first guess in Newton's iteration formula, we find the next guess to be ...

  x ≈ -0.746756

Iterating another time gives a solution accurate to 12 significant digits. The above result is well within the allowed tolerance of 10^-5.

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Of course, you know that the Newton's Method iterator for finding the "next guess" is ...

  next guess = x - f(x)/f'(x)

where f'(x) is the derivative of f(x), and the value of x is the current guess.

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Comment on this solution

For the purposes of Newton's method a numerical estimate of the derivative of the function is sufficient. Most graphing calculators have a derivative function, so all you need to do is specify the function you want to evaluate the derivative of. You could go to the trouble to develop the exact formula for the derivative of this function, but it isn't necessary to obtain an answer that is close enough.

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We have shown two separate uses of the iteration function. In practice, this calculator (Desmos) gives the function result as you type the argument, so all you need to do is copy the function output to its input to obtain calculator accuracy (12 significant digits) in one pass. A calculator like a TI-84 can do a similar thing by making the first iteration, Y₂(-0.747), then repeatedly computing Y₂(Ans), that is, using the previous answer in the iteration function again. For that calculator, one additional iteration is all that is needed to get to its 10 significant digit limit.