# Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 49

$x=1$, local minima at $x=1$

#### Work Step by Step

To find the critical point, we put $f'(x)=0$, so we have $$f(x)=\frac{e^{x}}{x}\Longrightarrow f'(x)=\frac{xe^x-e^x}{x^2}=0,$$ then $xe^{x}=e^x$ and hence $x=1$. So the critical point is $x=1$. Moreover, we have $$f''(x)=\frac{x^2(xe^x)-2x(xe^x-e^x)}{x^4} =\frac{x^3-2x(x-1)}{x^4}e^x$$ then $$f''(1)=e>0$$ then $f(x)$ has a local minima at $x=1$.

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