Question

find the critical values of $g(x)=\frac{9}{x^{2}+4x - 7}$. the critical values are: $x=$

Explanation:

Step1: Recall critical - value definition

Critical values of a function \(y = g(x)\) occur where \(g^{\prime}(x)=0\) or \(g^{\prime}(x)\) is undefined. First, use the quotient rule. If \(g(x)=\frac{u(x)}{v(x)}\), then \(g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}\). Here, \(u(x) = 9\) and \(v(x)=x^{2}+4x - 7\).

Step2: Calculate derivatives of \(u(x)\) and \(v(x)\)

\(u^{\prime}(x) = 0\) since the derivative of a constant is 0, and \(v^{\prime}(x)=2x + 4\) using the power - rule \((x^n)^\prime=nx^{n - 1}\).

Step3: Apply the quotient rule

\(g^{\prime}(x)=\frac{0\times(x^{2}+4x - 7)-9\times(2x + 4)}{(x^{2}+4x - 7)^2}=\frac{-9(2x + 4)}{(x^{2}+4x - 7)^2}\).

Step4: Find where \(g^{\prime}(x) = 0\) and where it is undefined

Set \(g^{\prime}(x)=0\). Then \(\frac{-9(2x + 4)}{(x^{2}+4x - 7)^2}=0\). The numerator must be 0 for the fraction to be 0. So, \(-9(2x + 4)=0\), which gives \(2x+4 = 0\), and \(x=-2\). The derivative is undefined when the denominator \((x^{2}+4x - 7)^2 = 0\), i.e., when \(x^{2}+4x - 7=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\), here \(a = 1\), \(b = 4\), \(c=-7\). Then \(x=\frac{-4\pm\sqrt{4^{2}-4\times1\times(-7)}}{2\times1}=\frac{-4\pm\sqrt{16 + 28}}{2}=\frac{-4\pm\sqrt{44}}{2}=\frac{-4\pm2\sqrt{11}}{2}=-2\pm\sqrt{11}\).

Answer:

\(x=-2,x=-2+\sqrt{11},x=-2 - \sqrt{11}\)