The interval in which the function $f(x)=x^x, x>0$, is strictly increasing is
Solution
<p>First note that </p>
<p>$f(x)=x^x=e^{\,x\ln x}\,,\quad x>0.$ </p>
<p>Differentiating gives </p>
<p>$f'(x)=x^x\bigl(\ln x+1\bigr)\,.$ </p>
<p>Since $x^x>0$, the sign of $f'(x)$ is the sign of $\ln x+1$. Hence </p>
<p>$ f'(x)>0\quad\Longleftrightarrow\quad \ln x>-1\quad\Longleftrightarrow\quad x>\frac1e. $ </p>
<p>So $f$ is strictly increasing for all $x>1/e$. Among the given choices that corresponds to </p>
<p>Option D: $\displaystyle\bigl[\tfrac1e,\infty\bigr)$.</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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