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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} - 8 x + 15}{x - 5}, & x < 5 \\ a x^{2} + b x + 2, & 5 \leq x < 7 \\ - a + b + 2 x, & x \geq 7 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=5\) and \(\displaystyle x=7\) must be equal. This gives the first equation at \(\displaystyle x = 5\) as \(\displaystyle 2 = 25 a + 5 b + 2\) and second equation at \(\displaystyle x = 7\) as \(\displaystyle 49 a + 7 b + 2 = - a + b + 14\). Getting each equation into standard form gives the system \(\displaystyle 25 a + 5 b = 0 \quad 50 a + 6 b = 12\). Solving the system gives \(\displaystyle a=\frac{3}{5}\) and \(\displaystyle b=-3\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} - 8 x + 15}{x - 5}, & x < 5 \\ a x^{2} + b x + 2, & 5 \leq x < 7 \\ - a + b + 2 x, & x \geq 7 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=5$ and $x=7$ must be equal. This gives the first equation at $x = 5$ as $2 = 25 a + 5 b + 2$ and second equation at $x = 7$ as $49 a + 7 b + 2 = - a + b + 14$. Getting each equation into standard form gives the system $25 a + 5 b = 0 \quad 50 a + 6 b = 12$. Solving the system gives $a=\frac{3}{5}$ and $b=-3$.}
\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} - 8 x + 15}{x - 5}, & x < 5 \\ a x^{2} + b x + 2, & 5 \leq x < 7 \\ - a + b + 2 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-%208%20x%20%2B%2015%7D%7Bx%20-%205%7D%2C%20%26%20x%20%3C%205%20%5C%5C%20a%20x%5E%7B2%7D%20%2B%20b%20x%20%2B%202%2C%20%26%20%205%20%5Cleq%20x%20%3C%207%20%5C%5C%20-%20a%20%2B%20b%20%2B%202%20x%2C%20%26%20%20x%20%5Cgeq%207%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} - 8 x + 15}{x - 5}, & x < 5 \\ a x^{2} + b x + 2, & 5 \leq x < 7 \\ - a + b + 2 x, & x \geq 7 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} - 8 x + 15}{x - 5}, & x < 5 \\ a x^{2} + b x + 2, & 5 \leq x < 7 \\ - a + b + 2 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=5 " src="/equation_images/%20%5Cdisplaystyle%20x%3D5%20" alt="LaTeX: \displaystyle x=5 " data-equation-content=" \displaystyle x=5 " /> 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 = 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%205%20" alt="LaTeX: \displaystyle x = 5 " data-equation-content=" \displaystyle x = 5 " /> as <img class="equation_image" title=" \displaystyle 2 = 25 a + 5 b + 2 " src="/equation_images/%20%5Cdisplaystyle%202%20%3D%2025%20a%20%2B%205%20b%20%2B%202%20" alt="LaTeX: \displaystyle 2 = 25 a + 5 b + 2 " data-equation-content=" \displaystyle 2 = 25 a + 5 b + 2 " /> 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 + 2 = - a + b + 14 " src="/equation_images/%20%5Cdisplaystyle%2049%20a%20%2B%207%20b%20%2B%202%20%3D%20-%20a%20%2B%20b%20%2B%2014%20" alt="LaTeX: \displaystyle 49 a + 7 b + 2 = - a + b + 14 " data-equation-content=" \displaystyle 49 a + 7 b + 2 = - a + b + 14 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle 25 a + 5 b = 0 \quad 50 a + 6 b = 12 " src="/equation_images/%20%5Cdisplaystyle%2025%20a%20%2B%205%20b%20%3D%200%20%5Cquad%2050%20a%20%2B%206%20b%20%3D%2012%20" alt="LaTeX: \displaystyle 25 a + 5 b = 0 \quad 50 a + 6 b = 12 " data-equation-content=" \displaystyle 25 a + 5 b = 0 \quad 50 a + 6 b = 12 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=\frac{3}{5} " src="/equation_images/%20%5Cdisplaystyle%20a%3D%5Cfrac%7B3%7D%7B5%7D%20" alt="LaTeX: \displaystyle a=\frac{3}{5} " data-equation-content=" \displaystyle a=\frac{3}{5} " /> and <img class="equation_image" title=" \displaystyle b=-3 " src="/equation_images/%20%5Cdisplaystyle%20b%3D-3%20" alt="LaTeX: \displaystyle b=-3 " data-equation-content=" \displaystyle b=-3 " /> .</p> </p>