1. The problem is to study the function $$f(x) = \frac{e^x}{|x-1|}$$. 2. We will analyze the domain, intercepts, behavior near critical points, and limits. 3. The domain is all real numbers except where the denominator is zero, so $$x \neq 1$$. 4. The function is undefined at $$x=1$$ because of division by zero. 5. To find intercepts: - For the y-intercept, set $$x=0$$: $$f(0) = \frac{e^0}{|0-1|} = \frac{1}{1} = 1$$. - For x-intercepts, solve $$f(x)=0$$, which requires $$e^x=0$$, impossible, so no x-intercepts. 6. Analyze behavior near $$x=1$$: - As $$x \to 1^+$$, denominator $$|x-1| = x-1 \to 0^+$$, numerator $$e^x \to e^1 = e$$, so $$f(x) \to +\infty$$. - As $$x \to 1^-$$, denominator $$|x-1| = 1-x \to 0^+$$, numerator $$e^x \to e$$, so $$f(x) \to +\infty$$. 7. The function has a vertical asymptote at $$x=1$$. 8. For large $$x$$: - As $$x \to +\infty$$, $$e^x$$ grows exponentially, denominator $$|x-1|$$ grows linearly, so $$f(x) \to +\infty$$. - As $$x \to -\infty$$, $$e^x \to 0$$, denominator $$|x-1| \to \infty$$, so $$f(x) \to 0^+$$. 9. To find critical points, differentiate: $$f(x) = \frac{e^x}{|x-1|}$$. For $$x>1$$, $$|x-1| = x-1$$, so $$f(x) = \frac{e^x}{x-1}$$. Derivative: $$f'(x) = \frac{e^x (x-1) - e^x \cdot 1}{(x-1)^2} = \frac{e^x (x-2)}{(x-1)^2}$$. Set numerator zero: $$e^x (x-2) = 0 \implies x=2$$. For $$x<1$$, $$|x-1| = 1 - x$$, so $$f(x) = \frac{e^x}{1-x}$$. Derivative: $$f'(x) = \frac{e^x (1-x) - e^x (-1)}{(1-x)^2} = \frac{e^x (2 - x)}{(1-x)^2}$$. Set numerator zero: $$2 - x = 0 \implies x=2$$, but $$x=2$$ is not in $$x<1$$, so no critical points there. 10. Therefore, the only critical point is at $$x=2$$. 11. Evaluate $$f(2)$$: $$f(2) = \frac{e^2}{|2-1|} = e^2$$. 12. Since $$f'(x)$$ changes sign around $$x=2$$ (positive to positive, denominator squared always positive), check sign: - For $$x>2$$, numerator $$e^x (x-2) > 0$$, so $$f'(x) > 0$$. - For $$1 < x < 2$$, numerator $$e^x (x-2) < 0$$, so $$f'(x) < 0$$. So $$f(x)$$ decreases on $$(1,2)$$ and increases on $$(2, \infty)$$, making $$x=2$$ a local minimum. 13. Summary: - Domain: $$\mathbb{R} \setminus \{1\}$$ - Vertical asymptote at $$x=1$$ - No x-intercepts - Y-intercept at $$(0,1)$$ - Local minimum at $$x=2$$ with value $$e^2$$ - $$f(x) \to 0^+$$ as $$x \to -\infty$$ - $$f(x) \to +\infty$$ as $$x \to 1^{\pm}$$ and $$x \to +\infty$$ Final answer: The function has a vertical asymptote at $$x=1$$, a local minimum at $$x=2$$ with value $$e^2$$, and no zeros.