Question #338262

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|


1
Expert's answer
2022-05-10T18:26:49-0400

f(x)=sin(x+π2)f(x)=sin{(x+\frac{\pi}{2})}

f(0)=sin(0+π2)=sinπ2=1f(0)=sin{(0+\frac{\pi}{2})}=sin{\frac{\pi}{2}}=1

f(π)=sin(π+π2)=sin3π2=1f(\pi)=sin{(\pi+\frac{\pi}{2})}=sin{\frac{3\pi}{2}}=-1

f(π2)=sin(π2+π2)=sinπ=0f(\frac{\pi}{2})=sin{(\frac{\pi}{2}+\frac{\pi}{2})}=sin{\pi}=0

f(x)=lnx1f(x)=ln|x-1|

f(0)=ln01=ln1=0f(0)=ln|0-1|=ln1=0

f(π)=lnπ1=0,76155f(\pi)=ln|\pi-1|=0,76155

f(π2)=lnπ21=0,56072f(\frac{\pi}{2})=ln|{\frac{\pi}{2}}-1|=-0,56072

f(x)=sin(x+π2)=cos(x+π2)f'(x)=sin'(x+\frac{\pi}{2})=cos(x+\frac{\pi}{2})

f(x)=cos(x+π2)=0f’(x)= cos(x+\frac{\pi}{2})=0

x+π2=π2+πk,kZx+\frac{\pi}{2}=\frac{\pi}{2}+\pi k, k\in Z

x=π2π2+πk,kZx=\frac{\pi}{2}-\frac{\pi}{2}+\pi k, k\in Z

x=πk,kZx=\pi k,k \in Z

f(x)=(lnx1)=1x1f’(x)=(ln|x-1|)’=\frac{1}{|x-1|}

1x1=0\frac{1}{|x-1|}=0

xR/(1)x\in R/(1)



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