Answer :
Answer:
x = -1.964636
Step-by-step explanation:
Given equation;
eˣ = 4 - x²
This can be re-written as;
eˣ - 4 + x² = 0
Let
f(x) = eˣ - 4 + x² -----------(i)
To use Newton's method, we need to get the first derivative of the above equation as follows;
f¹(x) = eˣ - 0 + 2x
f¹(x) = eˣ + 2x -----------(ii)
The graph of f(x) has been attached to this response.
As shown in the graph, the curve intersects the x-axis twice - around x = -2 and x = 1. These are the approximate roots of the equation.
Since the question requires that we use the negative root, then we start using the Newton's law with a guess of x₀ = -2 at n=0
From Newton's method,
[tex]x_{n+1} = x_n + \frac{f(x_{n})}{f^1(x_{n})}[/tex]
=> When n=0, the equation becomes;
[tex]x_{1} = x_0 - \frac{f(x_{0})}{f^1(x_{0})}[/tex]
[tex]x_{1} = -2 - \frac{f(-2)}{f^1(-2)}[/tex]
Where f(-2) and f¹(-2) are found by plugging x = -2 into equations (i) and (ii) as follows;
f(-2) = e⁻² - 4 + (-2)²
f(-2) = e⁻² = 0.13533528323
And;
f¹(2) = e⁻² + 2(-2)
f¹(2) = e⁻² - 4 = -3.8646647167
Therefore
[tex]x_{1} = -2 - \frac{0.13533528323}{-3.8646647167}[/tex]
[tex]x_{1} = -2 - \frac{0.13533528323}{-3.8646647167}[/tex]
[tex]x_{1} = -2 - -0.03501863503[/tex]
[tex]x_{1} = -2 + 0.03501863503[/tex]
[tex]x_{1} = -1.9649813649[/tex]
[tex]x_{1} = -1.96498136[/tex] [to 8 decimal places]
=> When n=1, the equation becomes;
[tex]x_{2} = x_1 - \frac{f(x_{1})}{f^1(x_{1})}[/tex]
[tex]x_{2} = -1.96498136 - \frac{f(-1.9649813)}{f^1(-1.9649813)}[/tex]
Following the same procedure as above we have
[tex]x_{2} = -1.96463563[/tex]
=> When n=2, the equation becomes;
[tex]x_{3} = x_2 - \frac{f(x_{2})}{f^1(x_{2})}[/tex]
[tex]x_{3} = -1.96463563- \frac{f( -1.96463563)}{f^1( -1.96463563)}[/tex]
Following the same procedure as above we have
[tex]x_{3} = -1.96463560[/tex]
From the values of [tex]x_2[/tex] and [tex]x_3[/tex], it can be seen that there is no change in the first 6 decimal places, therefore, it is safe to say that the value of the negative root of the equation is approximately -1.964636 to 6 decimal places.
Newton's method of approximation is one of the several ways of estimating values.
The approximated value of [tex]\mathbf{e^x = 4 - x^2}[/tex] to 6 decimal places is [tex]\mathbf{ -1.964636}[/tex]
The equation is given as:
[tex]\mathbf{e^x = 4 - x^2}[/tex]
Equate to 0
[tex]\mathbf{4 - x^2 = 0}[/tex]
So, we have:
[tex]\mathbf{x^2 = 4}[/tex]
Take square roots of both sides
[tex]\mathbf{ x= \pm 2}[/tex]
So, the negative root is:
[tex]\mathbf{x = -2}[/tex]
[tex]\mathbf{e^x = 4 - x^2}[/tex] becomes [tex]\mathbf{f(x) = e^x - 4 + x^2 }[/tex]
Differentiate
[tex]\mathbf{f'(x) = e^x +2x }[/tex]
Using Newton's method of approximation, we have:
[tex]\mathbf{x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}}[/tex]
When x = -2, we have:
[tex]\mathbf{f'(-2) = e^{(-2)} +2(-2) = -3.86466471676}[/tex]
[tex]\mathbf{f(-2) = e^{-2} - 4 + (-2)^2 = 0.13533528323}[/tex]
So, we have:
[tex]\mathbf{x_{1} = -2 - \frac{0.13533528323}{-3.86466471676}}[/tex]
[tex]\mathbf{x_{1} = -2 + \frac{0.13533528323}{3.86466471676}}[/tex]
[tex]\mathbf{x_{1} = -1.96498136}[/tex]
Repeat the above process for repeated x values.
We have:
[tex]\mathbf{x_{2} = -1.96463563}[/tex]
[tex]\mathbf{x_{3} = -1.96463560}[/tex]
Up till the 6th decimal places,
[tex]\mathbf{x_2 = x_3}[/tex]
Hence, the approximated value of [tex]\mathbf{e^x = 4 - x^2}[/tex] to 6 decimal places is [tex]\mathbf{ -1.964636}[/tex]
Read more about Newton approximation at:
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