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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.
\(\displaystyle f(x) = \begin{cases} \frac{x^{2} + 10 x + 21}{x + 3}, & x < -3 \\ a x^{2} + b x + 5, & -3 \leq x < -2 \\ - a + b + 2 x, & x \geq -2 \end{cases}\)
To be continuous the two sided limits at \(\displaystyle x=-3\) and \(\displaystyle x=-2\) must be equal. This gives the first equation at \(\displaystyle x = -3\) as \(\displaystyle 4 = 9 a - 3 b + 5\) and second equation at \(\displaystyle x = -2\) as \(\displaystyle 4 a - 2 b + 5 = - a + b - 4\). Getting each equation into standard form gives the system \(\displaystyle 9 a - 3 b = -1 \quad 5 a - 3 b = -9\). Solving the system gives \(\displaystyle a=2\) and \(\displaystyle b=\frac{19}{3}\).
\begin{question}Find the values of $a$ and $b$ that make $f$ continuous everywhere.\newline$f(x) = \begin{cases} \frac{x^{2} + 10 x + 21}{x + 3}, & x < -3 \\ a x^{2} + b x + 5, & -3 \leq x < -2 \\ - a + b + 2 x, & x \geq -2 \end{cases}$
\soln{9cm}{To be continuous the two sided limits at $x=-3$ and $x=-2$ must be equal. This gives the first equation at $x = -3$ as $4 = 9 a - 3 b + 5$ and second equation at $x = -2$ as $4 a - 2 b + 5 = - a + b - 4$. Getting each equation into standard form gives the system $9 a - 3 b = -1 \quad 5 a - 3 b = -9$. Solving the system gives $a=2$ and $b=\frac{19}{3}$.}
\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} + 10 x + 21}{x + 3}, & x < -3 \\ a x^{2} + b x + 5, & -3 \leq x < -2 \\ - a + b + 2 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%2010%20x%20%2B%2021%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%205%2C%20%26%20%20-3%20%5Cleq%20x%20%3C%20-2%20%5C%5C%20-%20a%20%2B%20b%20%2B%202%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} + 10 x + 21}{x + 3}, & x < -3 \\ a x^{2} + b x + 5, & -3 \leq x < -2 \\ - a + b + 2 x, & x \geq -2 \end{cases} " data-equation-content=" \displaystyle f(x) = \begin{cases} \frac{x^{2} + 10 x + 21}{x + 3}, & x < -3 \\ a x^{2} + b x + 5, & -3 \leq x < -2 \\ - a + b + 2 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=-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=-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 = -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 + 5 " src="/equation_images/%20%5Cdisplaystyle%204%20%3D%209%20a%20-%203%20b%20%2B%205%20" alt="LaTeX: \displaystyle 4 = 9 a - 3 b + 5 " data-equation-content=" \displaystyle 4 = 9 a - 3 b + 5 " /> 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 + 5 = - a + b - 4 " src="/equation_images/%20%5Cdisplaystyle%204%20a%20-%202%20b%20%2B%205%20%3D%20-%20a%20%2B%20b%20-%204%20" alt="LaTeX: \displaystyle 4 a - 2 b + 5 = - a + b - 4 " data-equation-content=" \displaystyle 4 a - 2 b + 5 = - a + b - 4 " /> . Getting each equation into standard form gives the system <img class="equation_image" title=" \displaystyle 9 a - 3 b = -1 \quad 5 a - 3 b = -9 " src="/equation_images/%20%5Cdisplaystyle%209%20a%20-%203%20b%20%3D%20-1%20%5Cquad%205%20a%20-%203%20b%20%3D%20-9%20" alt="LaTeX: \displaystyle 9 a - 3 b = -1 \quad 5 a - 3 b = -9 " data-equation-content=" \displaystyle 9 a - 3 b = -1 \quad 5 a - 3 b = -9 " /> . Solving the system gives <img class="equation_image" title=" \displaystyle a=2 " src="/equation_images/%20%5Cdisplaystyle%20a%3D2%20" alt="LaTeX: \displaystyle a=2 " data-equation-content=" \displaystyle a=2 " /> and <img class="equation_image" title=" \displaystyle b=\frac{19}{3} " src="/equation_images/%20%5Cdisplaystyle%20b%3D%5Cfrac%7B19%7D%7B3%7D%20" alt="LaTeX: \displaystyle b=\frac{19}{3} " data-equation-content=" \displaystyle b=\frac{19}{3} " /> .</p> </p>