1. **State the problem:** We have the function $$f(x) = \frac{1}{1-x}$$ for $$0 \leq x < 1$$ and $$f(1) = 0$$. We want to explain why $$f$$ has a minimum value but no maximum value on the closed interval $$[0,1]$$. 2. **Analyze the function on the interval:** For $$0 \leq x < 1$$, the denominator $$1-x$$ is positive and decreases from 1 to 0 as $$x$$ approaches 1. 3. **Behavior near the endpoints:** - At $$x=0$$, $$f(0) = \frac{1}{1-0} = 1$$. - As $$x \to 1^-$$, $$1-x \to 0^+$$, so $$f(x) = \frac{1}{1-x} \to +\infty$$. - At $$x=1$$, $$f(1) = 0$$ by definition. 4. **Check continuity and values at endpoints:** - The function is continuous on $$[0,1)$$ but has a jump at $$x=1$$ where $$f(1) = 0$$. 5. **Minimum value:** - Since $$f(1) = 0$$ and for all $$x < 1$$, $$f(x) > 0$$, the minimum value on $$[0,1]$$ is $$0$$ at $$x=1$$. 6. **Maximum value:** - As $$x \to 1^-$$, $$f(x) \to +\infty$$, so there is no finite maximum value on $$[0,1]$$. **Final conclusion:** - $$f$$ attains its minimum value $$0$$ at $$x=1$$. - $$f$$ does not have a maximum value on $$[0,1]$$ because it grows without bound as $$x$$ approaches 1 from the left.