Answer:
x = [tex]e^{\frac{\pi}{2}}[/tex]
Step-by-step explanation:
Data provided in the question:
f'(x) = cos(ln x)
For finding points of maxima or minima , put f'(x) = 0
therefore,
cos(ln x) = 0
or
cos(ln x) = [tex]\cos(\frac{\pi}{2})[/tex]
or
ln x = [tex]\frac{\pi}{2}[/tex]
or
x = [tex]e^{\frac{\pi}{2}}[/tex]
now to check for maxima or minima at x = [tex]e^{\frac{\pi}{2}}[/tex]
f"(x) = [tex]-\sin(\ln x)\times\frac{1}{x}[/tex]
thus
at x = [tex]e^{\frac{\pi}{2}}[/tex]
f"(x) = [tex]-\sin(\ln (e^{\frac{\pi}{2}}))\times\frac{1}{e^{\frac{\pi}{2}}}[/tex]
= [tex]-\sin(e^{\frac{\pi}{2}})\times e^{-\frac{\pi}{2}}}[/tex]
= [tex]-1\times e^{-\frac{\pi}{2}}}[/tex] < 0
hence maxima at x = [tex]e^{\frac{\pi}{2}}[/tex]
