1. The first problem asks which statement cannot be used to conclude that $f(0)$ exists. - (A) $\lim_{x \to 0} f(x)$ exists means the limit exists but does not guarantee $f(0)$ is defined. - (B) $f$ is continuous at $x=0$ implies $f(0)$ exists and equals the limit. - (C) $f$ is differentiable at $x=0$ implies continuity and thus $f(0)$ exists. - (D) The graph has a y-intercept means $f(0)$ is defined. **Answer:** (A) cannot be used to conclude $f(0)$ exists. 2. The second problem involves the piecewise function: $$f(x) = \begin{cases} 2 & x < 5 \\ 2x - 4 & x \geq 5 \end{cases}$$ - Check continuity at $x=5$: $$\lim_{x \to 5^-} f(x) = 2$$ $$\lim_{x \to 5^+} f(x) = 2(5) - 4 = 6$$ Since limits from left and right differ, $f$ is not continuous at $x=5$. - Differentiability requires continuity, so $f$ is not differentiable at $x=5$. - The graph has a sharp corner at $x=5$ due to the jump in function values. **Answer:** (A) $f$ is not differentiable at $x=5$ because $f$ is not continuous at $x=5$. 3. The third problem asks which statement cannot be used to conclude $f$ is defined at $x=1$. - (A) $\lim_{x \to 1} f(x)$ exists does not guarantee $f(1)$ is defined. - (B) $f$ continuous at $x=1$ implies $f(1)$ exists. - (C) $f$ differentiable at $x=1$ implies continuity and $f(1)$ exists. - (D) The tangent line exists at $x=1$ implies differentiability and thus $f(1)$ exists. **Answer:** (A) cannot be used to conclude $f(1)$ is defined. 4. The fourth problem involves the piecewise function: $$f(x) = \begin{cases} x^2 - 20 & x < 5 \\ -x^2 + 20 & x \geq 5 \end{cases}$$ - Check continuity at $x=5$: $$\lim_{x \to 5^-} f(x) = 5^2 - 20 = 25 - 20 = 5$$ $$\lim_{x \to 5^+} f(x) = -25 + 20 = -5$$ Limits differ, so $f$ is not continuous at $x=5$. - Since $f$ is not continuous, it is not differentiable at $x=5$. - The graph has a sharp corner at $x=5$ due to the jump. - $f$ is defined at $x=5$ since the second piece applies. **Answer:** (A) $f$ is not differentiable at $x=5$ because $f$ is not continuous at $x=5$.