The value of the constant $c$ that makes the function $f$ defined by $f(x) = \begin{cases} x^2 - c^2, & \text{if } x < 4 \\ cx + 20, & \text{if } x \geq 4 \end{cases}$ continuous for all real numbers is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $-2$ ##
$\text{L.H.L.} = \lim\limits_{x \to 4^-} f(x) = \lim\limits_{h \to 0} f(4 - h)$
$= \lim\limits_{h \to 0} (4 - h)^2 - c^2 = (4 - 0)^2 - c^2 = 16 - c^2$
$\text{R.H.L.} = \lim\limits_{x \to 4^+} f(x) = \lim\limits_{h \to 0} f(4 + h)$
$= \lim\limits_{h \to 0} c(4 + h) + 20 = c(4 + 0) + 20 = 4c + 20$
Given function is continuous, then $\text{L.H.L.} = \text{R.H.L.}$
$16 - c^2 = 4c + 20 ⇒c^2 + 4c + 4 = 0$
$(c + 2)^2 = 0 ⇒c + 2 = 0 ⇒c = -2$