1. **Problem (a):** Sketch the function \( f(x) = \begin{cases} 2, & x \geq 0 \\ -1, & x < 0 \end{cases} \) and determine if it is continuous at \( x=0 \). - The function is constant 2 for \( x \geq 0 \) and constant -1 for \( x < 0 \). - To check continuity at \( x=0 \), check \( \lim_{x \to 0^-} f(x) = -1 \), \( \lim_{x \to 0^+} f(x) = 2 \), and \( f(0) = 2 \). - Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), the function is **not continuous** at \( x=0 \). 2. **Problem (b):** Find constants \( \alpha, \phi \) so that \( f(x) = \begin{cases} \alpha + \phi x, & x > 2 \\ 3, & x=2 \\ \phi - \alpha x^2, & x < 2 \end{cases} \) is continuous at \( x=2 \). - Continuity at \( x=2 \) requires: \[ \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x) = 3 \] - Left limit: \( \lim_{x \to 2^-} f(x) = \phi - \alpha (2)^2 = \phi - 4\alpha \) - Right limit: \( \lim_{x \to 2^+} f(x) = \alpha + 2\phi \) - Set equal to 3: \[ \phi - 4\alpha = 3 \] \[ \alpha + 2\phi = 3 \] - Solve system: From second: \( \alpha = 3 - 2\phi \) Substitute into first: \[ \phi - 4(3 - 2\phi) = 3 \implies \phi - 12 + 8\phi = 3 \implies 9\phi = 15 \implies \phi = \frac{15}{9} = \frac{5}{3} \] - Then \( \alpha = 3 - 2 \times \frac{5}{3} = 3 - \frac{10}{3} = \frac{9}{3} - \frac{10}{3} = -\frac{1}{3} \). 3. **Problem (c):** Discuss discontinuities. (i) \( f(x) = x^2 + x + 3 \) is a polynomial, continuous everywhere, so **no discontinuity**. (ii) \( f(x) = -\frac{x}{x-2} \) has a discontinuity at \( x=2 \) where denominator zero. - This is a **vertical asymptote (infinite discontinuity)**. (iii) \( f(x) = \begin{cases} x-2, & x<2 \\ x+2, & x \geq 2 \end{cases} \) - Check limits at \( x=2 \): Left limit: \( 2-2=0 \), right limit: \( 2+2=4 \), function value at 2 is \( 4 \). - Since limits differ, **jump discontinuity** at \( x=2 \). (iv) \( f(x) = \frac{x-3}{\sqrt{x} - \sqrt{3}} \) - At \( x=3 \), denominator zero. - Rationalize denominator: \[ f(x) = \frac{x-3}{\sqrt{x} - \sqrt{3}} \times \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}} = \frac{(x-3)(\sqrt{x} + \sqrt{3})}{x - 3} \] - Cancel \( x-3 \) (for \( x \neq 3 \)): \[ f(x) = \sqrt{x} + \sqrt{3} \] - So \( \lim_{x \to 3} f(x) = 2\sqrt{3} \). - Since \( f(3) \) undefined originally, this is a **removable discontinuity**. 4. **Problem (d)(i):** Show there exists \( c \in (0,1) \) such that \( f(c) = 0 \) for \( f(x) = x^3 - 3x + 1 \). - Evaluate \( f(0) = 1 > 0 \), \( f(1) = 1 - 3 + 1 = -1 < 0 \). - By Intermediate Value Theorem, since \( f \) continuous, there exists \( c \in (0,1) \) with \( f(c) = 0 \). 5. **Problem (d)(ii):** Compute fixed point of \( f(x) = x^2 - 3x + 4 \) in \( [1,3] \). - Fixed point satisfies \( f(x) = x \). - Solve \( x^2 - 3x + 4 = x \) \( \Rightarrow x^2 - 4x + 4 = 0 \). - Factor: \( (x-2)^2 = 0 \) so \( x=2 \). - Since \( 2 \in [1,3] \), fixed point is \( x=2 \). **Final answers:** (a) Not continuous at \( x=0 \). (b) \( \alpha = -\frac{1}{3}, \phi = \frac{5}{3} \). (c) (i) No discontinuity; (ii) Infinite discontinuity at \( x=2 \); (iii) Jump discontinuity at \( x=2 \); (iv) Removable discontinuity at \( x=3 \). (d)(i) Exists \( c \in (0,1) \) with \( f(c)=0 \). (d)(ii) Fixed point at \( x=2 \).