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})