Given f(x) = begin{cases} cx + 1, & x leq 3 | vedclass

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Question

(C) For a function $f(x)$ to be continuous at $x = a$,the left-hand limit,right-hand limit,and the value of the function at $x = a$ must be equal.Spec

Vedclass · Accepted Answer

$-\frac{2}{3}$

Vedclass · Answer

(C) For a function $f(x)$ to be continuous at $x = a$,the left-hand limit,right-hand limit,and the value of the function at $x = a$ must be equal.Specifically,$\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^+} f(x) = f(3)$.Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \ dx + 3, & x > 3 \end{cases}$.Left-hand limit at $x = 3$: $\lim_{x 	o 3^-} (cx + 1) = 3c + 1$.Right-hand limit at $x = 3$: $\lim_{x 	o 3^+} (dx + 3) = 3d + 3$.Since $f$ is continuous at $x = 3$,we have $3c + 1 = 3d + 3$.Rearranging the terms: $3c - 3d = 3 - 1$.$3(c - d) = 2$.$c - d = \frac{2}{3}$.We need to find $d - c$,so $d - c = -(c - d) = -\frac{2}{3}$.Therefore,the correct option is $C$.

Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \\ dx + 3, & x > 3 \end{cases}$. If $f$ is continuous at $x = 3$,then $d - c =$ . . . . . . .

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