hills function models how the amount of oxygen bound to hemoglobin in the blood depends on oxygen…

hills function models how the amount of oxygen bound to hemoglobin in the blood depends on oxygen concentration, p, in the surrounding tissues. in its most general form hills function models the fraction of hemoglobin molecules in blood that are bound to oxygen by using the following formula, where k is a positive constant, and n is a positive integer.\n\n$f(p)=\\frac{p^{n}}{k^{n}+p^{n}}$\n\n(a) calculate $f(p)$.\n(b) show that $f(p)>0$ for all $p>0$. this result means that increasing the oxygen concentration always increases the fraction of hemoglobin molecules that are bound to oxygen.\n\n(b) it is given that $p>0$, k is a positive constant, and n is a positive integer.\nsince n is always greater than 0, $k^{n}$ is always greater than 0, and $p^{n - 1}$ is always greater than 0, it follows that the numerator of $f(p)$ is always less than 0.

hills function models how the amount of oxygen bound to hemoglobin in the blood depends on oxygen concentration, p, in the surrounding tissues. in its most general form hills function models the fraction of hemoglobin molecules in blood that are bound to oxygen by using the following formula, where k is a positive constant, and n is a positive integer.\n\n$f(p)=\\frac{p^{n}}{k^{n}+p^{n}}$\n\n(a) calculate $f(p)$.\n(b) show that $f(p)>0$ for all $p>0$. this result means that increasing the oxygen concentration always increases the fraction of hemoglobin molecules that are bound to oxygen.\n\n(b) it is given that $p>0$, k is a positive constant, and n is a positive integer.\nsince n is always greater than 0, $k^{n}$ is always greater than 0, and $p^{n - 1}$ is always greater than 0, it follows that the numerator of $f(p)$ is always less than 0.

Answer

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y'=\frac{u'v - uv'}{v^{2}}). Here, (u = P^{n}), (u'=nP^{n - 1}), (v=k^{n}+P^{n}), and (v'=nP^{n - 1}). [ \begin{align*} f'(P)&=\frac{nP^{n - 1}(k^{n}+P^{n})-P^{n}(nP^{n - 1})}{(k^{n}+P^{n})^{2}}\ \end{align*} ]

Step2: Simplify the numerator

Expand the numerator: [ \begin{align*} nP^{n - 1}(k^{n}+P^{n})-P^{n}(nP^{n - 1})&=nP^{n - 1}k^{n}+nP^{2n - 1}-nP^{2n - 1}\ &=nP^{n - 1}k^{n} \end{align*} ] So, (f'(P)=\frac{nP^{n - 1}k^{n}}{(k^{n}+P^{n})^{2}})

Step3: Analyze the sign of (f'(P)) for (P>0)

Since (n) is a positive integer, (k>0), (P > 0). Then (nP^{n - 1}k^{n}>0) (because (n>0), (P^{n - 1}>0) for (P>0) and (n\geq1), (k^{n}>0)) and ((k^{n}+P^{n})^{2}>0) (sum of two positive terms squared)

Answer:

a. (f'(P)=\frac{nP^{n - 1}k^{n}}{(k^{n}+P^{n})^{2}})

b. Since (n>0), (k>0), (P > 0), (nP^{n - 1}k^{n}>0) and ((k^{n}+P^{n})^{2}>0), so (f'(P)=\frac{nP^{n - 1}k^{n}}{(k^{n}+P^{n})^{2}}>0) for all (P > 0)