Cook Harrington
10/23/2023 · Elementary School
[(a) La derivata della funzione \( f(x)=e^{x \log \left(1+\frac{1}{x}\right)} \) vale \( f^{\prime}(x)=e^{x \log \left(1+\frac{1}{x}\right)}\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right] \) ed è strettamente positiva per la disuguaglianza dell'esercizio 1.59 ; (b) la limitazione \( f(x)<e \) equivale a \( \log f(x)<\log e=1 \), cioè \[ \log \left(1+\frac{1}{x}\right)<\frac{1}{x} \] che è verificata in base all'esercizio 1.58 . La limitazione \( f(x)<e \), per \( x>0 \), segue anche dalla precedente parte (a) e dal fatto che \( f(x) \) converge ad \( e \), per \( x \rightarrow+\infty] \)
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(a) La derivata di \( f(x)=e^{x \log \left(1+\frac{1}{x}\right)} \) è \( f^{\prime}(x)=e^{x \log \left(1+\frac{1}{x}\right)}\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right] \) e è positiva.
(b) \( f(x)0 \), \( f(x)
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