Question

Show that f (x)=(1/x)ˣ has local maximum at x=1/e.

Answer 1

concept : any function y = f(x) has local maximum at x = a, if f'(a) = 0 and f"(a) < 0.

given function, f(x) = (1/x)ˣ = $e^{-xlnx}$

differentiate f(x) with respect to x,

f'(x) = $\frac{d(e^{-xlnx})}{dx}$

= $e^{-xlnx}\left(-x\times\frac{1}{x}-lnx\right)$

= (1/x)ˣ (-1 - lnx)

at x = 1/e , f'(1/e) = (1/1/e)^(1/e){-1-ln(1/e)}

= (e)^{1/e}(-1 + lne)

= e^{1/e}(-1 + 1) [ as we know, lne = 1 ]

= 0

hence, f'(1/e) = 0.....(1)

again differentiate with respect to x,

f"(x) = (1/x)ˣ (-1/x) + (1/x)ˣ (-1 - lnx)²

at x = 1/e, f"(1/e) = (1/1/e)^{1/e}(-1/1/e) + (1/1/e)^{1/e}{-1-ln(1/e)}²

= e^{1/e} × (-e) + (e)^{1/e}(-1 + 1)²

= - e^{1/e + 1} + 0 < 0

hence, f"(1/e) < 0 ....(2)

from equations (1) and (2),

at x = 1/e, f'(1/e) = 0 and f"(1/e) < 0

hence, f(x) has local maximum at x = 1/e