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What will happen if the bisection method is used with the function f(x) = tan(x) and a) [3,4] b) [1,3] Attempt: Check the signs of the function: $f(x) = tan(x)$ a) f(3)*f(4) = -0.165 <0 => the root is between 3 and b) f(1)*f(3)= -0.222<0 => the root is between 1 and 3 , but this is obviously wrong. Please, help. |
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The bisection method is applicable when we wish to solve $f(x) = 0$ for $x \in \mathbb{R}$, where $$\color{red}{f \text{ is a continuous function defined on an interval } [a, b]}$$ and $f(a)$ and $f(b)$ have opposite signs. In your case, in the domain $[3,4]$ the function $\tan(x)$ is continuous and hence you can claim that there is a root in this domain and use bisection method. However, in the domain $[1,3]$, $\tan(x)$ is discontinuous at $\pi/2 \in (1.55,1.6)$ and hence the bisection method is not applicable in this interval. |
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$f(x) = \tan{x}$ has a pole at $\pi/2 \approx 1.57$, about which $f$ changes sign without crossing the $x$-axis. Thus bisection is not applicable within any bracketed interval containing $x=\pi/2$. |
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