Question

find the root of 4x-e^x=0 that lies between 2 and 3 by Newton-raphson method.​

Answer 1

Answer:

The root of the equation $4x-e^x=0$ lies between $(2,3)$ is $2.157$.

Step-by-step explanation:

Newton-Raphson method is one of the method for finding the roots of the equations.

Newton-Rapson formula,

$p_{n+1}=p_n-\frac{f(p_n)}{f'(p_n)}$, where $p_{n+1}$ = the next approximation

Step 1 of 3

Consider the given equation as follows:

$f(x)=4x-e^x$ on the interval $(2,3)$

Differentiate the function $f(x)$ with respect to $x$ as follows:

$\frac{d}{dx}( f(x))=\frac{d}{dx}(4x)-\frac{d}{dx}(e^x)$

⇒ $f'(x)=4-e^x$

Starting with the approximation of $p_0=2$, for two iterations of Newton's method.

Step 2 of 3

First iteration:

$p_{1}=p_0-\frac{f(p_0)}{f'(p_0)}$

    $=2-\frac{f(2)}{f'(2)}$

    $=2-\frac{4(2)-e^2}{4-e^2}$

Since the value of $e^2$ is $7.38$.

⇒ $p_1=2-\frac{8-7.32}{4-7.32}$

        $=2-\frac{0.68}{-3.38}$

        $=2.201$

Step 3 of 3

Second iteration:

$p_{2}=p_1-\frac{f(p_1)}{f'(p_1)}$

    $=2.201-\frac{f(2.201)}{f'(2.201)}$

    $=2.201-\frac{4(2.201)-e^{2.201}}{4-e^{2.201}}$

Since the value of $e^{2.201}$ is $9.03$.

⇒ $p_2=2.201-\frac{8.804-9.03}{4-9.03}$

        $=2.201-\frac{-0.226}{-5.03}$

        $=2.157$

Thus, the root of the equation $4x-e^x=0$ lies between $(2,3)$ is $2.157$.

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