For what value(s) of X does f(x)=(e^x)/(x) has a local maximum? There is no local maximum 0 (1)/(2) 2 1

## Step 1: Find the derivative of the function<br />### To find the local maximum or minimum, we need to find the critical points by taking the derivative of the function and setting it to zero. The derivative of $f(x) = \frac{e^x}{x}$ is given by the quotient rule: $f'(x) = \frac{xe^x - e^x}{x^2} = \frac{e^x(x-1)}{x^2}$.<br /><br />## Step 2: Find the critical points<br />### Set the derivative equal to zero and solve for x: $\frac{e^x(x-1)}{x^2} = 0$. This implies $e^x(x-1) = 0$. Since $e^x$ is never zero, we have $x-1=0$, so $x=1$.<br /><br />## Step 3: Determine if the critical point is a maximum or minimum<br />### To determine if the critical point $x=1$ is a local maximum or minimum, we can use the second derivative test. The second derivative is $f''(x) = \frac{x^3e^x - 2x^2e^x(x-1) - x^2e^x(x-1) + 2xe^x(x-1)}{x^4} = \frac{e^x(x^2 - 2x + 2)}{x^3}$. Evaluating the second derivative at $x=1$, we get $f''(1) = \frac{e^1(1-2+2)}{1} = e > 0$. Since the second derivative is positive at $x=1$, the function has a local minimum at $x=1$.