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Questions: Algebra BusinessCalculus
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Find the values of \(\displaystyle a\) and \(\displaystyle b\) that make \(\displaystyle f\) continuous everywhere.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} - 6 x + 5}{x - 1}, & x < 1 \\ a x^{2} + b x + 3, & 1 \leq x < 3 \\ - a + b + 2 x, & x \geq 3 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=1\) and \(\displaystyle x=3\) must be equal. This gives the first equation at \(\displaystyle x = 1\) as \(\displaystyle -4 = a + b + 3\) and second equation at \(\displaystyle x = 3\) as \(\displaystyle 9 a + 3 b + 3 = - a + b + 6\). Getting each equation into standard form gives the system \(\displaystyle a + b = -7 \quad 10 a + 2 b = 3\). Solving the system gives \(\displaystyle a=\frac{17}{8}\) and \(\displaystyle b=- \frac{73}{8}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} - 6 x + 5}{x - 1}, & x < 1 \\ a x^{2} + b x + 3, & 1 \leq x < 3 \\ - a + b + 2 x, & x \geq 3 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=1$ and $x=3$ must be equal. This gives the first equation at $x = 1$ as $-4 = a + b + 3$ and second equation at $x = 3$ as $9 a + 3 b + 3 = - a + b + 6$. Getting each equation into standard form gives the system $a + b = -7 \quad 10 a + 2 b = 3$. Solving the system gives $a=\frac{17}{8}$ and $b=- \frac{73}{8}$.}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the values of <img class="equation_image" title=" \displaystyle a " src="/equation_images/%20%5Cdisplaystyle%20a%20" alt="LaTeX: \displaystyle a " data-equation-content=" \displaystyle a " /> and <img class="equation_image" title=" \displaystyle b " src="/equation_images/%20%5Cdisplaystyle%20b%20" alt="LaTeX: \displaystyle b " data-equation-content=" \displaystyle b " /> that make <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX: \displaystyle f " data-equation-content=" \displaystyle f " /> continuous everywhere.<br> <img class="equation_image" title=" \displaystyle f(x) = \begin{cases} \frac{x^{2} - 6 x + 5}{x - 1}, & x < 1 \\ a x^{2} + b x + 3, & 1 \leq x < 3 \\ - a + b + 2 x, & x \geq 3 \end{cases} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cbegin%7Bcases%7D%20%5Cfrac%7Bx%5E%7B2%7D%20-%206%20x%20%2B%205%7D%7Bx%20-%201%7D%2C%20%26%20x%20%3C%201%20%5C%5C%20a%20x%5E%7B2%7D%20%2B%20b%20x%20%2B%203%2C%20%26%20%201%20%5Cleq%20x%20%3C%203%20%5C%5C%20-%20a%20%2B%20b%20%2B%202%20x%2C%20%26%20%20x%20%5Cgeq%203%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} - 6 x + 5}{x - 1}, & x < 1 \\ a x^{2} + b x + 3, & 1 \leq x < 3 \\ - a + b + 2 x, & x \geq 3 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} - 6 x + 5}{x - 1}, & x < 1 \\ a x^{2} + b x + 3, & 1 \leq x < 3 \\ - a + b + 2 x, & x \geq 3 \end{cases} " /> </p> </p><p> <p>To be continuous the two sided limits at <img class="equation_image" title=" \displaystyle x=1 " src="/equation_images/%20%5Cdisplaystyle%20x%3D1%20" alt="LaTeX: \displaystyle x=1 " data-equation-content=" \displaystyle x=1 " /> and <img class="equation_image" title=" \displaystyle x=3 " src="/equation_images/%20%5Cdisplaystyle%20x%3D3%20" alt="LaTeX: \displaystyle x=3 " data-equation-content=" \displaystyle x=3 " /> must be equal. This gives the first equation at <img class="equation_image" title=" \displaystyle x = 1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%201%20" alt="LaTeX: \displaystyle x = 1 " data-equation-content=" \displaystyle x = 1 " /> as <img class="equation_image" title=" \displaystyle -4 = a + b + 3 " src="/equation_images/%20%5Cdisplaystyle%20-4%20%3D%20a%20%2B%20b%20%2B%203%20" alt="LaTeX: \displaystyle -4 = a + b + 3 " data-equation-content=" \displaystyle -4 = a + b + 3 " /> and second equation at <img class="equation_image" title=" \displaystyle x = 3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%203%20" alt="LaTeX: \displaystyle x = 3 " data-equation-content=" \displaystyle x = 3 " /> as <img class="equation_image" title=" \displaystyle 9 a + 3 b + 3 = - a + b + 6 " src="/equation_images/%20%5Cdisplaystyle%209%20a%20%2B%203%20b%20%2B%203%20%3D%20-%20a%20%2B%20b%20%2B%206%20" alt="LaTeX: \displaystyle 9 a + 3 b + 3 = - a + b + 6 " data-equation-content=" \displaystyle 9 a + 3 b + 3 = - a + b + 6 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle a + b = -7 \quad 10 a + 2 b = 3 " src="/equation_images/%20%5Cdisplaystyle%20a%20%2B%20b%20%3D%20-7%20%5Cquad%2010%20a%20%2B%202%20b%20%3D%203%20" alt="LaTeX: \displaystyle a + b = -7 \quad 10 a + 2 b = 3 " data-equation-content=" \displaystyle a + b = -7 \quad 10 a + 2 b = 3 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=\frac{17}{8} " src="/equation_images/%20%5Cdisplaystyle%20a%3D%5Cfrac%7B17%7D%7B8%7D%20" alt="LaTeX: \displaystyle a=\frac{17}{8} " data-equation-content=" \displaystyle a=\frac{17}{8} " /> and <img class="equation_image" title=" \displaystyle b=- \frac{73}{8} " src="/equation_images/%20%5Cdisplaystyle%20b%3D-%20%5Cfrac%7B73%7D%7B8%7D%20" alt="LaTeX: \displaystyle b=- \frac{73}{8} " data-equation-content=" \displaystyle b=- \frac{73}{8} " /> .</p> </p>