Question

Prove f(x) = xˣ, x ∈ R⁺ is increasing if x > 1/e and decreasing if 0 < x < 1/e.

Answer 1

function, f(x) = $x^x$

⇒ f(x) = $e^{xlnx}$

differentiating both sides,

f'(x) = $e^{xlnx}(1+lnx)$

case 1 : x > 1/e

taking log base e both sides,

lnx > ln(1/e)

⇒lnx > ln(e^-1)

⇒lnx > -1ln(e)

[as we know, ln(e) = 1]

so, lnx > -1

⇒1 + lnx > 1 - 1

⇒1 + lnx > 0

and we know, exponential function is a positive function.so, $e^{xlnx}$ >0

hence, f'(x) = $e^{xlnx}(1+lnx)$ > 0.

therefore, function f(x) is increasing on x > 1/e

case 2 : 0 < x < 1/e

taking log base e

⇒ln(0) < lnx < ln(1/e)

⇒-∞ < lnx < -1

⇒-∞ < 1 + lnx < 0

i.e., (1 + lnx) < 0 and $e^{xlnx}$ > 0

so, f'(x) = $e^{xlnx}(1+lnx)$ < 0

hence, function f(x) is decreasing on 0 < x < 1/e .