Question

the function f(x) = \frac{6.5x^{2}-84.2x + 3740}{x^{2}+576} models the ph level, f(x), of a mouth x minutes after eating food containing sugar. the graph of this function is shown to the right. complete parts (a) through (e). a. use the graph to obtain a reasonable estimate, to the nearest tenth, of the ph level of the human mouth 30 minutes after a person eats food containing sugar. the ph level after 30 minutes is 4.8 (round to the nearest tenth as needed.) b. after eating sugar, when is the ph level the lowest? use the functions equation to determine the ph level at this time. the ph level is the lowest after minutes.

Explanation:

Step1: Recall the method to find the minimum of a rational - function

For a rational function \(y = \frac{f(x)}{g(x)}\), we can use calculus (find the derivative) or for a quadratic - like rational function, we can rewrite it in a more convenient form. First, the given function is \(f(x)=\frac{6.5x^{2}-84.2x + 3740}{x^{2}+576}\). We can rewrite it as \(f(x)=\frac{6.5x^{2}-84.2x + 3740}{x^{2}+576}=6.5-\frac{84.2x+64}{x^{2}+576}\). To find the minimum, we find its derivative using the quotient rule. The quotient rule states that if \(y=\frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = 6.5x^{2}-84.2x + 3740\), \(u^\prime=13x-84.2\), \(v=x^{2}+576\), and \(v^\prime = 2x\). Then \(f^\prime(x)=\frac{(13x - 84.2)(x^{2}+576)-(6.5x^{2}-84.2x + 3740)\times(2x)}{(x^{2}+576)^{2}}\).

Step2: Set the derivative equal to zero

Set \(f^\prime(x) = 0\). Since the denominator \((x^{2}+576)^{2}\gt0\) for all real \(x\), we only need to consider the numerator \((13x - 84.2)(x^{2}+576)-(6.5x^{2}-84.2x + 3740)\times(2x)=0\). Expand the left - hand side:
\[

$$\begin{align*} 13x^{3}+7488x-84.2x^{2}-48499.2-(13x^{3}-168.4x^{2}+7480x)&=0\\ 13x^{3}+7488x-84.2x^{2}-48499.2 - 13x^{3}+168.4x^{2}-7480x&=0\\ 84.2x^{2}+8x - 48499.2&=0 \end{align*}$$

\]
This is a quadratic equation \(ax^{2}+bx + c = 0\) where \(a = 84.2\), \(b = 8\), and \(c=-48499.2\).

Step3: Solve the quadratic equation

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 84.2\), \(b = 8\), and \(c=-48499.2\) into the formula:
\[

$$\begin{align*} x&=\frac{-8\pm\sqrt{8^{2}-4\times84.2\times(-48499.2)}}{2\times84.2}\\ &=\frac{-8\pm\sqrt{64 + 4\times84.2\times48499.2}}{168.4}\\ &=\frac{-8\pm\sqrt{64+1629731.84}}{168.4}\\ &=\frac{-8\pm\sqrt{1629795.84}}{168.4}\\ &=\frac{-8\pm1276.6}{168.4} \end{align*}$$

\]
We get two solutions for \(x\): \(x_1=\frac{-8 + 1276.6}{168.4}=\frac{1268.6}{168.4}\approx7.5\) and \(x_2=\frac{-8 - 1276.6}{168.4}\) (we discard the negative solution since \(x\) represents time in minutes after eating).

Step4: Find the minimum value of the function

Substitute \(x = 7.5\) into the function \(f(x)=\frac{6.5x^{2}-84.2x + 3740}{x^{2}+576}\):
\[

$$\begin{align*} f(7.5)&=\frac{6.5\times(7.5)^{2}-84.2\times7.5 + 3740}{(7.5)^{2}+576}\\ &=\frac{6.5\times56.25-631.5 + 3740}{56.25+576}\\ &=\frac{365.625-631.5 + 3740}{632.25}\\ &=\frac{3474.125}{632.25}\\ &\approx5.5 \end{align*}$$

\]

Answer:

The pH level is the lowest after \(7.5\) minutes.