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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} - 2 x - 8}{x - 4}, & x < 4 \\ a x^{2} + b x + 5, & 4 \leq x < 7 \\ - a + b + 3 x, & x \geq 7 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=4\) and \(\displaystyle x=7\) must be equal. This gives the first equation at \(\displaystyle x = 4\) as \(\displaystyle 6 = 16 a + 4 b + 5\) and second equation at \(\displaystyle x = 7\) as \(\displaystyle 49 a + 7 b + 5 = - a + b + 21\). Getting each equation into standard form gives the system \(\displaystyle 16 a + 4 b = 1 \quad 50 a + 6 b = 16\). Solving the system gives \(\displaystyle a=\frac{29}{52}\) and \(\displaystyle b=- \frac{103}{52}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} - 2 x - 8}{x - 4}, & x < 4 \\ a x^{2} + b x + 5, & 4 \leq x < 7 \\ - a + b + 3 x, & x \geq 7 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=4$ and $x=7$ must be equal. This gives the first equation at $x = 4$ as $6 = 16 a + 4 b + 5$ and second equation at $x = 7$ as $49 a + 7 b + 5 = - a + b + 21$. Getting each equation into standard form gives the system $16 a + 4 b = 1 \quad 50 a + 6 b = 16$. Solving the system gives $a=\frac{29}{52}$ and $b=- \frac{103}{52}$.}
\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} - 2 x - 8}{x - 4}, & x < 4 \\ a x^{2} + b x + 5, & 4 \leq x < 7 \\ - a + b + 3 x, & x \geq 7 \end{cases} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cbegin%7Bcases%7D%20%5Cfrac%7Bx%5E%7B2%7D%20-%202%20x%20-%208%7D%7Bx%20-%204%7D%2C%20%26%20x%20%3C%204%20%5C%5C%20a%20x%5E%7B2%7D%20%2B%20b%20x%20%2B%205%2C%20%26%20%204%20%5Cleq%20x%20%3C%207%20%5C%5C%20-%20a%20%2B%20b%20%2B%203%20x%2C%20%26%20%20x%20%5Cgeq%207%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} - 2 x - 8}{x - 4}, & x < 4 \\ a x^{2} + b x + 5, & 4 \leq x < 7 \\ - a + b + 3 x, & x \geq 7 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} - 2 x - 8}{x - 4}, & x < 4 \\ a x^{2} + b x + 5, & 4 \leq x < 7 \\ - a + b + 3 x, & x \geq 7 \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%3D4%20" alt="LaTeX: \displaystyle x=4 " data-equation-content=" \displaystyle x=4 " /> and <img class="equation_image" title=" \displaystyle x=7 " src="/equation_images/%20%5Cdisplaystyle%20x%3D7%20" alt="LaTeX: \displaystyle x=7 " data-equation-content=" \displaystyle x=7 " /> 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%204%20" alt="LaTeX: \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " /> as <img class="equation_image" title=" \displaystyle 6 = 16 a + 4 b + 5 " src="/equation_images/%20%5Cdisplaystyle%206%20%3D%2016%20a%20%2B%204%20b%20%2B%205%20" alt="LaTeX: \displaystyle 6 = 16 a + 4 b + 5 " data-equation-content=" \displaystyle 6 = 16 a + 4 b + 5 " /> and second equation at <img class="equation_image" title=" \displaystyle x = 7 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207%20" alt="LaTeX: \displaystyle x = 7 " data-equation-content=" \displaystyle x = 7 " /> as <img class="equation_image" title=" \displaystyle 49 a + 7 b + 5 = - a + b + 21 " src="/equation_images/%20%5Cdisplaystyle%2049%20a%20%2B%207%20b%20%2B%205%20%3D%20-%20a%20%2B%20b%20%2B%2021%20" alt="LaTeX: \displaystyle 49 a + 7 b + 5 = - a + b + 21 " data-equation-content=" \displaystyle 49 a + 7 b + 5 = - a + b + 21 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle 16 a + 4 b = 1 \quad 50 a + 6 b = 16 " src="/equation_images/%20%5Cdisplaystyle%2016%20a%20%2B%204%20b%20%3D%201%20%5Cquad%2050%20a%20%2B%206%20b%20%3D%2016%20" alt="LaTeX: \displaystyle 16 a + 4 b = 1 \quad 50 a + 6 b = 16 " data-equation-content=" \displaystyle 16 a + 4 b = 1 \quad 50 a + 6 b = 16 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=\frac{29}{52} " src="/equation_images/%20%5Cdisplaystyle%20a%3D%5Cfrac%7B29%7D%7B52%7D%20" alt="LaTeX: \displaystyle a=\frac{29}{52} " data-equation-content=" \displaystyle a=\frac{29}{52} " /> and <img class="equation_image" title=" \displaystyle b=- \frac{103}{52} " src="/equation_images/%20%5Cdisplaystyle%20b%3D-%20%5Cfrac%7B103%7D%7B52%7D%20" alt="LaTeX: \displaystyle b=- \frac{103}{52} " data-equation-content=" \displaystyle b=- \frac{103}{52} " /> .</p> </p>