If f(x)= x-4 |x-4| +a, &if x 4 is continuous at x = 4. Find a, b.

Given, f(x) is continuous at x = 4 & f(4) = a + b For f(x) to be continuous at x = 4, f(4)^- = f(4)^+= f(4) L.H.L=f(4)^-= _ x 4 x-4 |x-4| +a _ h 0 (4-h)-4 |(4-h)-4| +a _ h 0 (4-h-4) |(4-h-4)| +a _ h 0 (-h) |(-h)| +a _ h 0 (-h) h +a -1 L.H.L=f(4)^+= _ x 0 x-4 |x-4| _ h 0 (4+h)-4 |(4+h)-4| _ h 0 4+h-4 |4+h-4| _ h 0 h |h| _ h 0 1 |1| 1 Since, f(x) is is continuous at x = 4 & f(4) = a + b f(4)^-=f(4)^+=f(4) a-1=a+b=1 -1=1 =2 +b=1 =1-2 =-1

If f(x)= x-4 |x-4| +a, &if x <4 +b,&if x=4 x-4 |x-4| +b, & if x>4…

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Question

If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-4}{\text{|x}-4|}+\text{a}, &\text{if x} <4\\\text{a}+\text{b},&\text{if x}=4\\\frac{\text{x}-4}{\text{|x}-4|}+\text{b}, & \text{if x}>4\end{cases}$ is continuous at $x = 4$. Find $a, b$.

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Answer

Given,
f(x) is continuous at x = 4 & f(4) = a + b
For f(x) to be continuous at $x = 4, f(4)^- = f(4)^+= f(4)$
$\text{L.H.L}=\text{f(4)}^-=\lim\limits_{\text{x} \rightarrow 4}\frac{\text{x}-4}{|\text{x}-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h})-4}{|(4-\text{h})-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h}-4)}{|(4-\text{h}-4)|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{|(-\text{h})|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{\text{h}}+\text{a}$
$\Rightarrow\text{a}-1$
$\text{L.H.L}=\text{f(4)}^+=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}-4}{|\text{x}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4+\text{h})-4}{|(4+\text{h})-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{4+\text{h}-4}{|4+\text{h}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{|\text{h}|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{1}{|1|}$
$\Rightarrow1$
Since, f(x) is is continuous at x = 4 & f(4) = a + b
$\text{f(4)}^-=\text{f(4)}^+=\text{f(4)}$
$\therefore\ \text{a}-1=\text{a}+\text{b}=1$
$\Rightarrow\text{a}-1=1$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{a}+\text{b}=1$
$\Rightarrow\text{b}=1-2$
$\Rightarrow\text{b}=-1$

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