Find the values of a and b such that the func | vedclass

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(A) The given function $f$ is $f(x) = \begin{cases} 5, & 	ext{if } x \le 2 \ ax + b, & 	ext{if } 2 < x < 10 \ 21, & 	ext{if } x \ge 10 \end{cases

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$a=2, b=1$

Vedclass · Answer

(A) The given function $f$ is $f(x) = \begin{cases} 5, & 	ext{if } x \le 2 \ ax + b, & 	ext{if } 2 Since $f$ is a continuous function,it must be continuous at all points,including $x=2$ and $x=10$. For continuity at $x=2$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2)$ $5 = 2a + b = 5 \implies 2a + b = 5$  $(1)$ For continuity at $x=10$: $\lim_{x 	o 10^-} f(x) = \lim_{x 	o 10^+} f(x) = f(10)$ $10a + b = 21 = 21 \implies 10a + b = 21$  $(2)$ Subtracting equation $(1)$ from $(2)$: $(10a + b) - (2a + b) = 21 - 5$ $8a = 16 \implies a = 2$ Substituting $a=2$ into equation $(1)$: $2(2) + b = 5$ $4 + b = 5 \implies b = 1$ Thus,the values are $a=2$ and $b=1$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$ is a continuous function.

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