score on last try: 0.5 of 1 pts. see details for more. at least one scored part is incorrect. jump to first…

score on last try: 0.5 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. get a similar question you can retry this question below use newtons method to approximate a root of the equation e^0.7x^2 = 4 - x as follows. let x1 = 1 be the initial approximation. the second approximation x2 is 1.2586 and the third approximation x3is submit question

score on last try: 0.5 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. get a similar question you can retry this question below use newtons method to approximate a root of the equation e^0.7x^2 = 4 - x as follows. let x1 = 1 be the initial approximation. the second approximation x2 is 1.2586 and the third approximation x3is submit question

Answer

Explanation:

Step1: Define the function

Let $f(x)=e^{0.7x^{2}}+x - 4$. Then find its derivative $f'(x)=1.4xe^{0.7x^{2}}+1$.

Step2: Recall Newton's method formula

The formula for Newton's method is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$. We know $x_1 = 1$ and $x_2=1.2586$.

Step3: Calculate $f(x_2)$

$f(1.2586)=e^{0.7\times(1.2586)^{2}}+1.2586 - 4=e^{0.7\times1.584}+1.2586 - 4=e^{1.1088}+1.2586 - 4\approx3.031+1.2586 - 4=0.2896$.

Step4: Calculate $f'(x_2)$

$f'(1.2586)=1.4\times1.2586\times e^{0.7\times(1.2586)^{2}}+1=1.76204\times e^{1.1088}+1\approx1.76204\times3.031+1=5.34174+1 = 6.34174$.

Step5: Calculate $x_3$

$x_3=x_2-\frac{f(x_2)}{f'(x_2)}=1.2586-\frac{0.2896}{6.34174}\approx1.2586 - 0.0457=1.2129$.

Answer:

$1.2129$