Calculate a real root of the following function using bisection method | Numerical Method 2079

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Numerical Method20795 marksNumerical: bisection method root findingHalf-Interval method and Convergence

Calculate a real root of the following function using bisection method correct upto 3 significant figures. $x^2 - e^x = 3$ [5]

Equation: $x^2 - e^x = 3$ Rearranged as: $f(x) = x^2 - e^x - 3 = 0$ Required accuracy: 3 significant figures. --- $x$ $f(x) = x^2 - e^x - 3$ Sign --------- $-1$ $1 - 0.3679 - 3 = -2.368$ $-$ $-2$ $4 - 0.1353 - 3 = +0.865$ $+$ Sign change between $-2$ and $-1$ , so a root lies in $[-2, -1]$ . Let

Equation: $x^2 - e^x = 3$ Rearranged as: $f(x) = x^2 - e^x - 3 = 0$ Required accuracy: 3 significant figures. --- $x$ $f(x) = x^2 - e^x - 3$ Sign --------- $-1$ $1 - 0.3679 - 3 = -2.368$ $-$ $-2$ $4 - 0.1353 - 3 = +0.865$ $+$ Sign change between $-2$ and $-1$, so a root lies in $[-2, -1]$. Let

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Calculate a real root of the following function using bisection method correct upto 3 significant figures. $x^2 - e^x = 3$ [5]

Equation: $x^2 - e^x = 3$ Rearranged as: $f(x) = x^2 - e^x - 3 = 0$ Required accuracy: 3 significant figures. --- $x$ $f(x) = x^2 - e^x - 3$ Sign --------- $-1$ $1 - 0.3679 - 3 = -2.368$ $-$ $-2$ $4 - 0.1353 - 3 = +0.865$ $+$ Sign change between $-2$ and $-1$ , so a root lies in $[-2, -1]$ . Let

Calculate a real root of the following function using bisection method correct upto 3 significant figures. x2 - ex = 3