Show that, [tex]x^x[/tex] is strictly increasing over natural range, using exponent laws.

2 thoughts on “Show that, [tex]x^x[/tex] is strictly increasing over natural range, using exponent laws.”

1. Answer:

   Step-by-step explanation:

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2. Before we start.

   When the value of a function is always greater than the previous ones, it is called a strictly increasing function.

   Point of the Question.

   Over natural numbers, the next value of [tex]x[/tex] is [tex]x+1[/tex].

   Solution.

   Claim: [tex]f(x)=x^x[/tex] is strictly increasing over natural domain.

   Let one value in the domain be [tex]k[/tex], so its next value is [tex]k+1[/tex].

   First,

   [tex](k+1)^{k+1}=(k+1)^{k}\times (k+1)[/tex] …[I]

   Since [tex]k<k+1[/tex],

   [tex]k^k<(k+1)^{k}[/tex] …[II]

   According to [I] and [II],

   [tex]k^k<(k+1)^{k}<(k+1)^{k}\times (k+1)[/tex]

   [tex]\therefore k^k<(k+1)^{k+1}[/tex]

   [tex]\therefore \boxed{f(k)<f(k+1)}[/tex]

   So, it is strictly increasing for any value of k.

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