How do you use Newton's method to find the approximate solution to the equation #x+1/sqrtx=3#?

Answer 1
I got #x = 2.347296355#.

Newton's method (or the Newton-Raphson method) entails:

#x_(n ew) = x_(old) - (f(x_(old)))/(f'(x_(old))#
and this new guess, #x_(n ew)#, approaches the true answer from above, while #x_(old)# is any guess in the reals. Depending on the guess, it may not converge, but pick something reasonable.
First, I'd move all the #x# and constant terms to the left:
#x + 1/sqrtx - 3 = 0#

Then, taking the derivative gives the general form:

#f'(x) = 1 + (-1/2x^(-"3/2")) = 1 - 1/(2x^"3/2")#

So, if you write this in your TI-83+ calculator:

#"[insert guess]" -> x#
#x - "("x + 1"/"sqrtx - 3")/("1 - 1"/("2x"^"("3/2")"))" -> x#

which actually looks like this:

#[x_(old) - (x_(old) + 1/sqrt(x_(old)) - 3)/(1 - 1/(2(x_(old))^("3/2")))] -> x_(n ew)#
and then press enter several times, you should get the following output for an initial guess of #x = 4#:
#4 -> 2.4 -> 2.347433739 -> 2.347296356 -> 2.347296355#
so it converges to #color(blue)(2.347296355)#.

Check the solution:

#2.347296355 + 1/sqrt(2.347296355) stackrel(?)(=) 3#
#2.347296355 + 0.6527036447 = 3 = 3# #color(blue)(sqrt"")#
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Answer 2

To use Newton's method to find the approximate solution to the equation ( x + \frac{1}{\sqrt{x}} = 3 ), follow these steps:

  1. Start with an initial guess for the solution, ( x_0 ).
  2. Use the formula: [ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ] where ( f(x) ) is the given equation and ( f'(x) ) is its derivative.
  3. Repeat this process iteratively until the value of ( x ) converges to the desired accuracy.

In this specific equation, ( f(x) = x + \frac{1}{\sqrt{x}} - 3 ). To find ( f'(x) ), differentiate ( f(x) ) with respect to ( x ): [ f'(x) = 1 - \frac{1}{2x^{3/2}} ]

Now, choose an initial guess ( x_0 ) and use the formula to iteratively approximate the solution until convergence is achieved.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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