Partial Solutions


  1. There is no solution provided for this question.
  2. There are many possible solutions. The graph of one possible function, \(f(x)\), is shown below.
    Graph of constant function y=2 with an open point at (-1, 2).
    Since the given function satisfies \(f(x) = 2\) when \(x \neq -1\), we have \( \displaystyle \lim_{x \rightarrow -1^{-}} f(x) = \lim_{x \rightarrow -1^{+}} f(x) = 2\), but since \(f(x)\) is undefined at \(x=-1\), the function is discontinuous at \(x=-1\).
  3. There is no solution provided for this question.

    1. We have \( f(x) = \dfrac{x^2 + x - 2}{x^2 + 3x + 2} = \dfrac{(x - 1)(x + 2)}{(x + 2)(x + 1)} \) and so \( f(x) \) is undefined at \( x = -2 \) and \( x = -1 \), but continuous everywhere else.

      If \( x \neq -2 \), then \( f(x) = \dfrac{x - 1}{x + 1} \) and so \( \displaystyle \lim_{x \rightarrow -2} ~ f(x) = \lim_{x \to -2} ~ \dfrac{x - 1}{x + 1} = -1 \). Therefore, \( f \) has a removable discontinuity at \( x = -2 \) (which can be removed by defining \( f(-2) = -1 \)).

      If \( x \) is a number just to the right of \( -1 \), then \( f(x) = \dfrac{x - 1}{x + 1} \to \dfrac{-2}{0^+} \) and \( \displaystyle \lim_{x \to -1^+} ~ f(x) \to -\infty \). If \( x \) is a number just to the left of \( -1 \), then \( f(x) = \dfrac{x - 1}{x + 1} \to \dfrac{-2}{0^-} \) and \( \displaystyle \lim_{x \to -1^+} ~ f(x) \to \infty \). Thus \( f \) has an infinite discontinuity at \( x = -1 \).

    2. The domain of \( f(x) \) is \( \{x \mid x \in \mathbb{R} \} \). Since \( x + 2, x \) are continuous functions, \( f \) is continuous for all \( x \neq 3 \). Since \( \displaystyle \lim_{x \to 3^+} ~ f(x) = \lim_{x \to 3^+} ~ (x + 2) = 5 \) and \( \displaystyle \lim_{x \to 3^-} ~ f(x) = \lim_{x \to 3^-} ~ x = 3 \), \( \displaystyle \lim_{x \to 3} ~ f(x) \) does not exist, and \( f \) has a jump discontinuity at \( x = 3 \).

    3. Observe that \( f(2) \) is undefined and thus \( f \) is discontinuous at \( x = 2 \).

      If \( x \geq 2 \), then \( \left\lvert x - 2 \right\rvert = x - 2 \) and so \( f(x) = \dfrac{x + 1}{x - 2} \). So, \( f(x) \) is continuous for all \( x \gt 2 \). Also, \( \displaystyle \lim_{x \to 2^+} ~ f(x) = \lim_{x \to 2^+} ~ \dfrac{x + 1}{x - 2} \to \infty \).

      If \( x \leq 2 \), then \( \left\lvert x - 2 \right\rvert = -(x - 2) \) and so \( f(x) = \dfrac{x + 1}{-(x - 2)} = \dfrac{x + 1}{2 - x} \). So, \( f(x) \) is continuous for all \( x \lt 2 \). Also, \( \displaystyle \lim_{x \to 2^-} ~ f(x) = \lim_{x \to 2^-} ~ \dfrac{x + 1}{2 - x} \to \infty \).

      We conclude that \( x \) is discontinuous only at \( x = 2 \) and it is an infinite discontinuity.

  5. There is no solution provided for this question.
    1. No. Since \( \displaystyle \lim_{x \rightarrow 0^{+}} f(x) = \displaystyle \lim_{x \rightarrow 0^{+}} (x-2) = -2 \), \( \displaystyle \lim_{x \rightarrow 0^{-}} f(x) = \displaystyle \lim_{x \rightarrow 0^{+}} -(x+3) = -3 \) and \( f(0) = -2 \), \( f \) has a jump discontinuity at \( x = 0 \).
      Graph of piecewise function described in question; hole at (0,-3), jump discontinuity at x=0
    2. Yes, \( f \) is continuous for all \( x \). Since \( x+4 \) and \( x^{2}-2x \) are polynomials, \( f \) is continuous for all \( x \neq -1 \). Since \( \displaystyle \lim_{x \rightarrow -2^{+}} f(x) = \displaystyle \lim_{x \rightarrow -1^{+}} (x+4) = 3 \), \( \displaystyle \lim_{x \rightarrow -1^{-}} f(x) = \displaystyle \lim_{x \rightarrow -1^{+}}(x^{2}-2x) = (-1)^{2} +2 = 3 \) and \( f(-1) = 3 \), \( f \) is also continuous at \( x=-1 \).
      Graph of piecewise function described in question
  7. There is no solution provided for this question.
    1. By inspection, \( f(x) = x^2 - 4x \) for \( x \lt a \), which is continuous for all \( x \) since it is a polynomial function. Similarly, \( f(x) = -4 \) for \( x \gt a \), which is continuous for all \( x \) since it is a constant function. Then \( \displaystyle \lim_{x \rightarrow a^-} ~ f(x) = a^2 - 4a \) and \( \displaystyle \lim_{x \rightarrow a^+} ~ f(x) = -4 = f(a) \), so we need \( a^2 - 4a = -4 \) for \( f(x) \) to be continuous at \( x = a \). \begin{align*} a^2 - 4a &= -4 \\ a^2 - 4a + 4 &= 0 \\ (a - 2)^2 &= 0 \\ a &= 2 \end{align*} Therefore, \( f(x) \) is continuous for all \( x \) if and only if \( a = 2 \).

    2. Since \( a^2x + 2 \) and \( x^2 - bx + a \) are all continuous for any \( a \) and \( b \), we only need to check continuity at \( x = 3 \).

      We have \( f(3) = 5 \), \( \displaystyle \lim_{x \to 3^+} ~ f(x) = \lim_{x \to 3^+} ~ (a^2x + 2) = a^2(3) + 2 = 3a^2 + 2 \) and \( \displaystyle \lim_{x \to 3^-} ~ f(x) = \lim_{x \to 3^-} ~ (x^2 - bx + a) = (3)^2 - 3b + a = a - 3b + 9 \), so \( f \) is continuous if and only if \( 3a^2 + 2 = 5 ~~ (1) \) and \( a - 3b + 9 = 5 ~~ (2) \).

      From \( (1) \), \( 3a^2 = 3 \) and so \( a = \pm 1 \).

      From \( (2) \), \( a - 3b + 9 = 5 \).
      If \( a = 1 \), then \( 3b = 5 \) and \( b = \frac{5}{3} \).
      If \( a = -1 \), then \( 3b = 3 \) and \( b = 1 \).

      Therefore, the solutions for \( a \) and \( b \) are \( a = 1, b = \frac{5}{3} \) or \( a = -1, b = 1 \).

  9. There is no solution provided for this question.
    1. Any function that is undefined at \( x = 2 \) but whose limit exists at \( x = 2 \) will suffice. One possible solution is:
      Graph of y=x with a hole at (0,0)
    2. Any function that is undefined for \( x \notin [0, 1] \) with vertical asymptote at \( x = 1 \) will suffice. One possible solution is:
      Graph of function with endpoint at (0,1), increasing on (0,1), function approaches infinity as x approaches 1
    3. Any function that is undefined at \( x = -1, \frac{1}{2} \) but whose limit exists at \( x = -1 \) and has a vertical asymptote at \( x = \frac{1}{2} \) will suffice. One possible solution is:
      Graph of y=1/(x-0.5) + 1 with a hole at (-1,1/3)
    4. One possible solution is:
      Graph of y=3/(x - 1) for x<3, y=x for x>3, hole at (3,-1.5)
  11. There is no solution provided for this question.