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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} + x - 12}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -2 \\ - a + b + 3 x, & x \geq -2 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=-4\) and \(\displaystyle x=-2\) must be equal. This gives the first equation at \(\displaystyle x = -4\) as \(\displaystyle -7 = 16 a - 4 b + 2\) and second equation at \(\displaystyle x = -2\) as \(\displaystyle 4 a - 2 b + 2 = - a + b - 6\). Getting each equation into standard form gives the system \(\displaystyle 16 a - 4 b = -9 \quad 5 a - 3 b = -8\). Solving the system gives \(\displaystyle a=\frac{5}{28}\) and \(\displaystyle b=\frac{83}{28}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} + x - 12}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -2 \\ - a + b + 3 x, & x \geq -2 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=-4$ and $x=-2$ must be equal. This gives the first equation at $x = -4$ as $-7 = 16 a - 4 b + 2$ and second equation at $x = -2$ as $4 a - 2 b + 2 = - a + b - 6$. Getting each equation into standard form gives the system $16 a - 4 b = -9 \quad 5 a - 3 b = -8$. Solving the system gives $a=\frac{5}{28}$ and $b=\frac{83}{28}$.}
\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} + x - 12}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -2 \\ - a + b + 3 x, & x \geq -2 \end{cases} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cbegin%7Bcases%7D%20%5Cfrac%7Bx%5E%7B2%7D%20%2B%20x%20-%2012%7D%7Bx%20%2B%204%7D%2C%20%26%20x%20%3C%20-4%20%5C%5C%20a%20x%5E%7B2%7D%20%2B%20b%20x%20%2B%202%2C%20%26%20%20-4%20%5Cleq%20x%20%3C%20-2%20%5C%5C%20-%20a%20%2B%20b%20%2B%203%20x%2C%20%26%20%20x%20%5Cgeq%20-2%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} + x - 12}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -2 \\ - a + b + 3 x, & x \geq -2 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} + x - 12}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -2 \\ - a + b + 3 x, & x \geq -2 \end{cases} " /> </p> </p><p> <p>To be continuous the two sided limits at <img class="equation_image" title=" \displaystyle x=-4 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-4%20" alt="LaTeX: \displaystyle x=-4 " data-equation-content=" \displaystyle x=-4 " /> and <img class="equation_image" title=" \displaystyle x=-2 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-2%20" alt="LaTeX: \displaystyle x=-2 " data-equation-content=" \displaystyle x=-2 " /> must be equal. This gives the first equation at <img class="equation_image" title=" \displaystyle x = -4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-4%20" alt="LaTeX: \displaystyle x = -4 " data-equation-content=" \displaystyle x = -4 " /> as <img class="equation_image" title=" \displaystyle -7 = 16 a - 4 b + 2 " src="/equation_images/%20%5Cdisplaystyle%20-7%20%3D%2016%20a%20-%204%20b%20%2B%202%20" alt="LaTeX: \displaystyle -7 = 16 a - 4 b + 2 " data-equation-content=" \displaystyle -7 = 16 a - 4 b + 2 " /> and second equation at <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX: \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> as <img class="equation_image" title=" \displaystyle 4 a - 2 b + 2 = - a + b - 6 " src="/equation_images/%20%5Cdisplaystyle%204%20a%20-%202%20b%20%2B%202%20%3D%20-%20a%20%2B%20b%20-%206%20" alt="LaTeX: \displaystyle 4 a - 2 b + 2 = - a + b - 6 " data-equation-content=" \displaystyle 4 a - 2 b + 2 = - a + b - 6 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle 16 a - 4 b = -9 \quad 5 a - 3 b = -8 " src="/equation_images/%20%5Cdisplaystyle%2016%20a%20-%204%20b%20%3D%20-9%20%5Cquad%205%20a%20-%203%20b%20%3D%20-8%20" alt="LaTeX: \displaystyle 16 a - 4 b = -9 \quad 5 a - 3 b = -8 " data-equation-content=" \displaystyle 16 a - 4 b = -9 \quad 5 a - 3 b = -8 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=\frac{5}{28} " src="/equation_images/%20%5Cdisplaystyle%20a%3D%5Cfrac%7B5%7D%7B28%7D%20" alt="LaTeX: \displaystyle a=\frac{5}{28} " data-equation-content=" \displaystyle a=\frac{5}{28} " /> and <img class="equation_image" title=" \displaystyle b=\frac{83}{28} " src="/equation_images/%20%5Cdisplaystyle%20b%3D%5Cfrac%7B83%7D%7B28%7D%20" alt="LaTeX: \displaystyle b=\frac{83}{28} " data-equation-content=" \displaystyle b=\frac{83}{28} " /> .</p> </p>