Question

Consider the function \begin{equation} f(x) = \begin{cases} 4 \cdot x + 8 & x < -\frac{1}{4} \\ 8 & x = -\frac{1}{4} \\ 112 \cdot x^2 & x > -\frac{1}{4} \end{cases} \end{equation} Tick all of the following statements that are correct. $\lim_{x \to -\frac{1}{4}} f(x) = f(-\frac{1}{4}).$ $\lim_{x \to (-\frac{1}{4})^+} f(x) = 8.$ $f$ has a jump discontinuity at $x = -\frac{1}{4}.$ $\lim_{x \to (-\frac{1}{4})^-} f(x) = 7.$ $f$ is continuous at $x = -\frac{1}{4}.$ $f$ is discontinuous at $x = -\frac{1}{4}.$ $\lim_{x \to -\frac{1}{4}} f(x)$ exists. $f$ has a removable discontinuity at $x = -\frac{1}{4}.$

          Consider the function
\begin{equation*} f(x) = \begin{cases} 4 \cdot x + 8 & x < -\frac{1}{4} \\ 8 & x = -\frac{1}{4} \\ 112 \cdot x^2 & x > -\frac{1}{4} \end{cases} \end{equation*}
Tick all of the following statements that are correct.
$\lim_{x \to -\frac{1}{4}} f(x) = f(-\frac{1}{4}).$
$\lim_{x \to (-\frac{1}{4})^+} f(x) = 8.$
$f$ has a jump discontinuity at $x = -\frac{1}{4}.$
$\lim_{x \to (-\frac{1}{4})^-} f(x) = 7.$
$f$ is continuous at $x = -\frac{1}{4}.$
$f$ is discontinuous at $x = -\frac{1}{4}.$
$\lim_{x \to -\frac{1}{4}} f(x)$ exists.
$f$ has a removable discontinuity at $x = -\frac{1}{4}.$

Consider the function

f(x) = 4 · x + 8 x < -(1)/(4)
8 x = -(1)/(4)
112 · x^2 x > -(1)/(4)

Tick all of the following statements that are correct.
limx → -(1)/(4) f(x) = f(-(1)/(4)).
limx → (-(1)/(4))^+ f(x) = 8.
f has a jump discontinuity at x = -(1)/(4).
limx → (-(1)/(4))^- f(x) = 7.
f is continuous at x = -(1)/(4).
f is discontinuous at x = -(1)/(4).
limx → -(1)/(4) f(x) exists.
f has a removable discontinuity at x = -(1)/(4).

Added by Juan M.

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