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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} + 2 x - 3}{x + 3}, & x < -3 \\ a x^{2} + b x + 1, & -3 \leq x < -1 \\ - a + b + 2 x, & x \geq -1 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=-3\) and \(\displaystyle x=-1\) must be equal. This gives the first equation at \(\displaystyle x = -3\) as \(\displaystyle -4 = 9 a - 3 b + 1\) and second equation at \(\displaystyle x = -1\) as \(\displaystyle a - b + 1 = - a + b - 2\). Getting each equation into standard form gives the system \(\displaystyle 9 a - 3 b = -5 \quad 2 a - 2 b = -3\). Solving the system gives \(\displaystyle a=- \frac{1}{12}\) and \(\displaystyle b=\frac{17}{12}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} + 2 x - 3}{x + 3}, & x < -3 \\ a x^{2} + b x + 1, & -3 \leq x < -1 \\ - a + b + 2 x, & x \geq -1 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=-3$ and $x=-1$ must be equal. This gives the first equation at $x = -3$ as $-4 = 9 a - 3 b + 1$ and second equation at $x = -1$ as $a - b + 1 = - a + b - 2$. Getting each equation into standard form gives the system $9 a - 3 b = -5 \quad 2 a - 2 b = -3$. Solving the system gives $a=- \frac{1}{12}$ and $b=\frac{17}{12}$.}
\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 - 3}{x + 3}, & x < -3 \\ a x^{2} + b x + 1, & -3 \leq x < -1 \\ - a + b + 2 x, & x \geq -1 \end{cases} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cbegin%7Bcases%7D%20%5Cfrac%7Bx%5E%7B2%7D%20%2B%202%20x%20-%203%7D%7Bx%20%2B%203%7D%2C%20%26%20x%20%3C%20-3%20%5C%5C%20a%20x%5E%7B2%7D%20%2B%20b%20x%20%2B%201%2C%20%26%20%20-3%20%5Cleq%20x%20%3C%20-1%20%5C%5C%20-%20a%20%2B%20b%20%2B%202%20x%2C%20%26%20%20x%20%5Cgeq%20-1%20%5Cend%7Bcases%7D%20" alt="LaTeX: \displaystyle f(x) = \begin{cases} \frac{x^{2} + 2 x - 3}{x + 3}, & x < -3 \\ a x^{2} + b x + 1, & -3 \leq x < -1 \\ - a + b + 2 x, & x \geq -1 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} + 2 x - 3}{x + 3}, & x < -3 \\ a x^{2} + b x + 1, & -3 \leq x < -1 \\ - a + b + 2 x, & x \geq -1 \end{cases} " /> </p> </p><p> <p>To be continuous the two sided limits at <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 " /> and <img class="equation_image" title=" \displaystyle x=-1 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-1%20" alt="LaTeX: \displaystyle x=-1 " data-equation-content=" \displaystyle x=-1 " /> must be equal. This gives the first 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 -4 = 9 a - 3 b + 1 " src="/equation_images/%20%5Cdisplaystyle%20-4%20%3D%209%20a%20-%203%20b%20%2B%201%20" alt="LaTeX: \displaystyle -4 = 9 a - 3 b + 1 " data-equation-content=" \displaystyle -4 = 9 a - 3 b + 1 " /> and second equation at <img class="equation_image" title=" \displaystyle x = -1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1%20" alt="LaTeX: \displaystyle x = -1 " data-equation-content=" \displaystyle x = -1 " /> as <img class="equation_image" title=" \displaystyle a - b + 1 = - a + b - 2 " src="/equation_images/%20%5Cdisplaystyle%20a%20-%20b%20%2B%201%20%3D%20-%20a%20%2B%20b%20-%202%20" alt="LaTeX: \displaystyle a - b + 1 = - a + b - 2 " data-equation-content=" \displaystyle a - b + 1 = - a + b - 2 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle 9 a - 3 b = -5 \quad 2 a - 2 b = -3 " src="/equation_images/%20%5Cdisplaystyle%209%20a%20-%203%20b%20%3D%20-5%20%5Cquad%202%20a%20-%202%20b%20%3D%20-3%20" alt="LaTeX: \displaystyle 9 a - 3 b = -5 \quad 2 a - 2 b = -3 " data-equation-content=" \displaystyle 9 a - 3 b = -5 \quad 2 a - 2 b = -3 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=- \frac{1}{12} " src="/equation_images/%20%5Cdisplaystyle%20a%3D-%20%5Cfrac%7B1%7D%7B12%7D%20" alt="LaTeX: \displaystyle a=- \frac{1}{12} " data-equation-content=" \displaystyle a=- \frac{1}{12} " /> and <img class="equation_image" title=" \displaystyle b=\frac{17}{12} " src="/equation_images/%20%5Cdisplaystyle%20b%3D%5Cfrac%7B17%7D%7B12%7D%20" alt="LaTeX: \displaystyle b=\frac{17}{12} " data-equation-content=" \displaystyle b=\frac{17}{12} " /> .</p> </p>