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Solve for a and b so that f(x) is continuous at all points.


f(x) = \begin{cases}(x^2-4)/(x-2) & \text{if } x\ \textless \ 2 \\ax^2-bx+3 & \text{if } 2\le x \ \textless \ 3\\2x-a+b & \text{if } x\ge 3\end{cases}

1 Answer

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Answer:


a=(1)/(2),\;\;b=(1)/(2)

Explanation:

A function f is continuous at x = a when:


\bullet\quad\textsf{$f(a)$\;is\;d\:\!efined.}


\bullet\quad\textsf{$\displaystyle \lim_(x \to a) f(x)$\;exists.}


\bullet\quad\displaystyle \lim_(x \to a) f(x) = f(a)

To ensure that f(x) is continuous at x = 2 and x = 3, we need to make sure that the limit of f(x) as x approaches 2 from the left is equal to the limit of f(x) as x approaches 2 from the right, and similarly for x = 3.

First, factor the numerator and simplify the rational function:


f(x)=(x^2-4)/(x-2)=((x+2)(x-2))/(x-2)=x+2

As x approaches 2 from the left, x < 2. Therefore, to find the limit as x approaches 2 from the left, substitute x = 2 into the first sub-function:


\displaystyle \lim_(x \to 2^(-)) f(x)=2+2=4

As x approaches 2 from the right, x > 2. Therefore, to find the limit as x approaches 2 from the right, substitute x = 2 into the second sub-function:


\begin{aligned}\displaystyle \lim_(x \to 2^(+)) f(x)&amp;=a(2)^2-b(2)+3\\&amp;=4a-2b+3\end{aligned}

To ensure continuity at x = 2, equate the limits:


4=4a - 2b + 3

Solve for b in terms of a:


\begin{aligned}4&amp;=4a - 2b + 3&amp;\\ 2b+4&amp;=4a+3\\2b&amp;=4a-1\\b&amp;=2a-(1)/(2)\end{aligned}

Now, we need to find a and b so that f(x) is continuous at x = 3.

As x approaches 3 from the left, x < 3. Therefore, to find the limit as x approaches 3 from the left, substitute x = 3 into the second sub-function:


\begin{aligned}\displaystyle \lim_(x \to 3^(-)) f(x)&amp;=a(3)^2 - b(3) + 3\\&amp;= 9a - 3b + 3\end{aligned}

As x approaches 3 from the right, x > 3. Therefore, to find the limit as x approaches 3 from the right, substitute x = 3 into the third sub-function:


\begin{aligned}\displaystyle \lim_(x \to 3^(+)) f(x)&amp;= 2(3) - a + b\\&amp;= 6 - a + b\end{aligned}

To ensure continuity at x = 3, equate the limits:


9a - 3b + 3 = 6-a+b

Solve for b in terms of a:


\begin{aligned}9a - 3b + 3 &amp;= 6-a+b\\10a-3b+3&amp;=6+b\\10a+3&amp;=6+4b\\10a-3&amp;=4b\\4b&amp;=10a-3\\b&amp;=(5)/(2)a-(3)/(4)\end{aligned}

Substitute this expression for b into the equation obtained from continuity at x = 2:


(5)/(2)a-(3)/(4)=2a-(1)/(2)

Solve for a:


\begin{aligned}(5)/(2)a-(3)/(4)&amp;=2a-(1)/(2)\\\\(1)/(2)a-(3)/(4)&amp;=-(1)/(2)\\\\(1)/(2)a&amp;=(1)/(4)\\\\a&amp;=(1)/(2) \end{aligned}

Substitute the found value of a into the expression for b and solve for b:


\begin{aligned}b&amp;=2\left((1)/(2)\right)-(1)/(2)\\b&amp;=1-(1)/(2)\\b&amp;=(1)/(2)\end{aligned}

Therefore, the values of a and b that make f(x) continuous at all points are:


a=(1)/(2),\;\;b=(1)/(2)

So the function f(x) is:


f(x)=\begin{cases} (x^2-4)/(x-2)&amp;\text{if}\;\;x < 2\\\\(1)/(2)x^2-(1)/(2)x+3\quad&amp;\text{if}\;\;2 \leq x < 3\\\\2x&amp;\text{if}\;\;x \geq 3\end{cases}

Solve for a and b so that f(x) is continuous at all points. f(x) = \begin{cases}(x-example-1
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