Monotonicity

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Question

$Use the function f (x) = x^{1 / x}, x > 0, to determine the bigger of the two$ $numbers e^{π} and π^{e} .$

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Solution

$\begin{array}{l} f (x) = x^{1 / x}, (x > 0) \\ \Rightarrow f^{'} (x) = x^{1 / x} \frac{(1 - 1 n x)}{x^{2}} = 0 \Rightarrow x = e \end{array}$
$\begin{array}{l} \Rightarrow f^{'} (x) \leq 0, when x \geq e \\ \Rightarrow f (x) decreases when x \geq e \\ And f^{'} (x) > 0, when 0 < x < e \\ \Rightarrow f (x) increases when 0 < x < e \\ Now π > e \\ \Rightarrow f (π) < f (e) (as f (x) decreasesin [e, \infty]) \\ \Rightarrow π^{1 / π} < e^{1 / e} \end{array}$
$Now raise positive power e π on both sides.$
$\begin{array}{ll} {(π^{1 / π})}^{e π} < {(e^{1 e})}^{e π} \\ s o, π^{e} > e^{π} \end{array}$

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Monotonicity

Remember concepts with our Masterclasses.

Use the function f(x)=x1/x,x>0, to determine the bigger of the two numbers eπ and πe.

f(x)=x1/x,(x>0)⇒ f′(x)=x1/x(1−1nx)x2=0⇒x=e⇒f′(x)≤0, when x≥e⇒f(x) decreases when x≥e And f′(x)>0, when 0<x<e⇒f(x) increases when 0<x<e Now π>e⇒f(π)<f(e)( as f(x) decreasesin [e,∞])⇒π1/π<e1/e Now raise positive power eπ on both sides. π1/πeπ<e1eeπso, πe>eπ

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$Use the function f (x) = x^{1 / x}, x > 0, to determine the bigger of the two$ $numbers e^{π} and π^{e} .$