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Consider the equation and the graph:
[tex] \frac{2}{x+4}=3^x+1 [/tex]

The approximate solution to the given equation after three iterations of successive approximations is when [tex] x [/tex] is about [tex] \square [/tex].


Sagot :

To solve the equation [tex]\(\frac{2}{x+4} = 3^x + 1\)[/tex] using successive approximations, we start with an initial guess for [tex]\(x\)[/tex], typically denoted as [tex]\(x_0\)[/tex]. After that, we iteratively refine this guess.

1. Initial guess [tex]\(x_0 = 0\)[/tex]:
- Start with [tex]\(x_0 = 0\)[/tex].

2. First iteration ([tex]\(x_1\)[/tex]):
- Compute [tex]\(f(x_0)\)[/tex] where [tex]\(f(x) = \frac{2}{x+4} - 3^x - 1\)[/tex].
- [tex]\(f(0) = \frac{2}{0+4} - 3^0 - 1 = \frac{2}{4} - 1 - 1 = 0.5 - 1 - 1 = -1.5\)[/tex].
- Update [tex]\(x_1 = f(x_0) + x_0 = -1.5 + 0 = -1.5\)[/tex].

3. Second iteration ([tex]\(x_2\)[/tex]):
- Compute [tex]\(f(x_1)\)[/tex].
- [tex]\(f(-1.5) = \frac{2}{-1.5+4} - 3^{-1.5} - 1 \approx \frac{2}{2.5} - 0.1924500897 - 1 \)[/tex].
- Update [tex]\(x_2 = f(x_1) + x_1 \approx -0.3924500897 + (-1.5) \approx -1.8924500897\)[/tex].

4. Third iteration ([tex]\(x_3\)[/tex]):
- Compute [tex]\(f(x_2)\)[/tex].
- [tex]\(f(-1.8924500897) = \frac{2}{-1.8924500897+4} - 3^{-1.8924500897} - 1 \approx \frac{2}{2.1075499103} - 0.1760773632 - 1 \)[/tex].
- Update [tex]\(x_3 = f(x_2) + x_2 \approx -0.1760773632 + (-1.8924500897) \approx -2.06852745296\)[/tex].

Thus, after three iterations of successive approximations, the solution to the equation [tex]\(\frac{2}{x+4} = 3^x + 1\)[/tex] is approximately when [tex]\(x \approx -2.06852745296\)[/tex].

So, the correct answer to the given equation after three iterations is when [tex]\(x\)[/tex] is about [tex]\(\boxed{-2.06852745296}\)[/tex].