Answer to Question #338262 in Calculus for mia

Question

Solve the corresponding equation for the appropriate interval by using the following two root finding techniques :

a) algebraic approach

sin(x+ π /2)=ln|x-1|

Answer 1

$f(x)=sin{(x+\frac{\pi}{2})}$

$f(0)=sin{(0+\frac{\pi}{2})}=sin{\frac{\pi}{2}}=1$

$f(\pi)=sin{(\pi+\frac{\pi}{2})}=sin{\frac{3\pi}{2}}=-1$

$f(\frac{\pi}{2})=sin{(\frac{\pi}{2}+\frac{\pi}{2})}=sin{\pi}=0$

$f(x)=ln|x-1|$

$f(0)=ln|0-1|=ln1=0$

$f(\pi)=ln|\pi-1|=0,76155$

$f(\frac{\pi}{2})=ln|{\frac{\pi}{2}}-1|=-0,56072$

$f'(x)=sin'(x+\frac{\pi}{2})=cos(x+\frac{\pi}{2})$

$f’(x)= cos(x+\frac{\pi}{2})=0$

$x+\frac{\pi}{2}=\frac{\pi}{2}+\pi k, k\in Z$

$x=\frac{\pi}{2}-\frac{\pi}{2}+\pi k, k\in Z$

$x=\pi k,k \in Z$

$f’(x)=(ln|x-1|)’=\frac{1}{|x-1|}$

$\frac{1}{|x-1|}=0$

$x\in R/(1)$