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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} + 11 x + 28}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -3 \\ - a + b + 2 x, & x \geq -3 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=-4\) and \(\displaystyle x=-3\) must be equal. This gives the first equation at \(\displaystyle x = -4\) as \(\displaystyle 3 = 16 a - 4 b + 2\) and second equation at \(\displaystyle x = -3\) as \(\displaystyle 9 a - 3 b + 2 = - a + b - 6\). Getting each equation into standard form gives the system \(\displaystyle 16 a - 4 b = 1 \quad 10 a - 4 b = -8\). Solving the system gives \(\displaystyle a=\frac{3}{2}\) and \(\displaystyle b=\frac{23}{4}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} + 11 x + 28}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -3 \\ - a + b + 2 x, & x \geq -3 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=-4$ and $x=-3$ must be equal. This gives the first equation at $x = -4$ as $3 = 16 a - 4 b + 2$ and second equation at $x = -3$ as $9 a - 3 b + 2 = - a + b - 6$. Getting each equation into standard form gives the system $16 a - 4 b = 1 \quad 10 a - 4 b = -8$. Solving the system gives $a=\frac{3}{2}$ and $b=\frac{23}{4}$.}
\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} + 11 x + 28}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \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%2B%2011%20x%20%2B%2028%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%20-3%20%5C%5C%20-%20a%20%2B%20b%20%2B%202%20x%2C%20%26%20%20x%20%5Cgeq%20-3%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} + 11 x + 28}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \leq x < -3 \\ - a + b + 2 x, & x \geq -3 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} + 11 x + 28}{x + 4}, & x < -4 \\ a x^{2} + b x + 2, & -4 \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=-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=-3 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-3%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 = -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 3 = 16 a - 4 b + 2 " src="/equation_images/%20%5Cdisplaystyle%203%20%3D%2016%20a%20-%204%20b%20%2B%202%20" alt="LaTeX: \displaystyle 3 = 16 a - 4 b + 2 " data-equation-content=" \displaystyle 3 = 16 a - 4 b + 2 " /> and second equation at <img class="equation_image" title=" \displaystyle x = -3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-3%20" alt="LaTeX: \displaystyle x = -3 " data-equation-content=" \displaystyle x = -3 " /> as <img class="equation_image" title=" \displaystyle 9 a - 3 b + 2 = - a + b - 6 " src="/equation_images/%20%5Cdisplaystyle%209%20a%20-%203%20b%20%2B%202%20%3D%20-%20a%20%2B%20b%20-%206%20" alt="LaTeX: \displaystyle 9 a - 3 b + 2 = - a + b - 6 " data-equation-content=" \displaystyle 9 a - 3 b + 2 = - a + b - 6 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle 16 a - 4 b = 1 \quad 10 a - 4 b = -8 " src="/equation_images/%20%5Cdisplaystyle%2016%20a%20-%204%20b%20%3D%201%20%5Cquad%2010%20a%20-%204%20b%20%3D%20-8%20" alt="LaTeX: \displaystyle 16 a - 4 b = 1 \quad 10 a - 4 b = -8 " data-equation-content=" \displaystyle 16 a - 4 b = 1 \quad 10 a - 4 b = -8 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=\frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20a%3D%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX: \displaystyle a=\frac{3}{2} " data-equation-content=" \displaystyle a=\frac{3}{2} " /> and <img class="equation_image" title=" \displaystyle b=\frac{23}{4} " src="/equation_images/%20%5Cdisplaystyle%20b%3D%5Cfrac%7B23%7D%7B4%7D%20" alt="LaTeX: \displaystyle b=\frac{23}{4} " data-equation-content=" \displaystyle b=\frac{23}{4} " /> .</p> </p>