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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} - 1}{x - 1}, & x < 1 \\ a x^{2} + b x + 1, & 1 \leq x < 4 \\ - a + b + 3 x, & x \geq 4 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=1\) and \(\displaystyle x=4\) must be equal. This gives the first equation at \(\displaystyle x = 1\) as \(\displaystyle 2 = a + b + 1\) and second equation at \(\displaystyle x = 4\) as \(\displaystyle 16 a + 4 b + 1 = - a + b + 12\). Getting each equation into standard form gives the system \(\displaystyle a + b = 1 \quad 17 a + 3 b = 11\). Solving the system gives \(\displaystyle a=\frac{4}{7}\) and \(\displaystyle b=\frac{3}{7}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} - 1}{x - 1}, & x < 1 \\ a x^{2} + b x + 1, & 1 \leq x < 4 \\ - a + b + 3 x, & x \geq 4 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=1$ and $x=4$ must be equal. This gives the first equation at $x = 1$ as $2 = a + b + 1$ and second equation at $x = 4$ as $16 a + 4 b + 1 = - a + b + 12$. Getting each equation into standard form gives the system $a + b = 1 \quad 17 a + 3 b = 11$. Solving the system gives $a=\frac{4}{7}$ and $b=\frac{3}{7}$.}
\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} - 1}{x - 1}, & x < 1 \\ a x^{2} + b x + 1, & 1 \leq x < 4 \\ - a + b + 3 x, & x \geq 4 \end{cases} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cbegin%7Bcases%7D%20%5Cfrac%7Bx%5E%7B2%7D%20-%201%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%201%2C%20%26%20%201%20%5Cleq%20x%20%3C%204%20%5C%5C%20-%20a%20%2B%20b%20%2B%203%20x%2C%20%26%20%20x%20%5Cgeq%204%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} - 1}{x - 1}, & x < 1 \\ a x^{2} + b x + 1, & 1 \leq x < 4 \\ - a + b + 3 x, & x \geq 4 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} - 1}{x - 1}, & x < 1 \\ a x^{2} + b x + 1, & 1 \leq x < 4 \\ - a + b + 3 x, & x \geq 4 \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=4 " src="/equation_images/%20%5Cdisplaystyle%20x%3D4%20" alt="LaTeX: \displaystyle x=4 " data-equation-content=" \displaystyle x=4 " /> 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 2 = a + b + 1 " src="/equation_images/%20%5Cdisplaystyle%202%20%3D%20a%20%2B%20b%20%2B%201%20" alt="LaTeX: \displaystyle 2 = a + b + 1 " data-equation-content=" \displaystyle 2 = a + b + 1 " /> and second equation at <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX: \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " /> as <img class="equation_image" title=" \displaystyle 16 a + 4 b + 1 = - a + b + 12 " src="/equation_images/%20%5Cdisplaystyle%2016%20a%20%2B%204%20b%20%2B%201%20%3D%20-%20a%20%2B%20b%20%2B%2012%20" alt="LaTeX: \displaystyle 16 a + 4 b + 1 = - a + b + 12 " data-equation-content=" \displaystyle 16 a + 4 b + 1 = - a + b + 12 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle a + b = 1 \quad 17 a + 3 b = 11 " src="/equation_images/%20%5Cdisplaystyle%20a%20%2B%20b%20%3D%201%20%5Cquad%2017%20a%20%2B%203%20b%20%3D%2011%20" alt="LaTeX: \displaystyle a + b = 1 \quad 17 a + 3 b = 11 " data-equation-content=" \displaystyle a + b = 1 \quad 17 a + 3 b = 11 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=\frac{4}{7} " src="/equation_images/%20%5Cdisplaystyle%20a%3D%5Cfrac%7B4%7D%7B7%7D%20" alt="LaTeX: \displaystyle a=\frac{4}{7} " data-equation-content=" \displaystyle a=\frac{4}{7} " /> and <img class="equation_image" title=" \displaystyle b=\frac{3}{7} " src="/equation_images/%20%5Cdisplaystyle%20b%3D%5Cfrac%7B3%7D%7B7%7D%20" alt="LaTeX: \displaystyle b=\frac{3}{7} " data-equation-content=" \displaystyle b=\frac{3}{7} " /> .</p> </p>