Calc-Ok Question
Tara’s heart rate during a workout is modeled by the differentiable function \(h\), where \(h(t)\) is measured in beats per minute and \(t\) is minutes from the start of the workout. Which expression gives Tara’s average heart rate from \(t=30\) to \(t=60\)?
(A) \(\displaystyle \int_{30}^{60} h(t)\,dt\)
(B) \(\displaystyle \frac{1}{30}\int_{30}^{60} h(t)\,dt\)
(C) \(\displaystyle \frac{1}{30}\int_{30}^{60} h'(t)\,dt\)
(D) \(\displaystyle \frac{h'(30)+h'(60)}{2}\)
(B) \(\displaystyle \frac{1}{30}\int_{30}^{60} h(t)\,dt\)
(C) \(\displaystyle \frac{1}{30}\int_{30}^{60} h'(t)\,dt\)
(D) \(\displaystyle \frac{h'(30)+h'(60)}{2}\)
▶️ Answer/Explanation
Detailed solution
The average value of a continuous function \(f\) on \([a,b]\) is \[ \frac{1}{b-a}\int_{a}^{b} f(x)\,dx. \] Here \(f=h\), \(a=30\), \(b=60\), so the average heart rate is \[ \frac{1}{60-30}\int_{30}^{60} h(t)\,dt=\frac{1}{30}\int_{30}^{60} h(t)\,dt. \] ✅ Answer: (B)
Calc-Ok Question
Let \(f\) be the function defined by \[ f(x)= \begin{cases} k^{3}+x, & x<3,\\[4pt] \dfrac{16}{k^{2}-x}, & x\ge 3, \end{cases} \] where \(k>0\). For what value of \(k\), if any, is \(f\) continuous?
(A) \(2.081\)
(B) \(2.646\)
(C) \(8.550\)
(D) There is no such value of \(k\)
(B) \(2.646\)
(C) \(8.550\)
(D) There is no such value of \(k\)
▶️ Answer/Explanation
Detailed solution
Continuity at \(x=3\) requires the left- and right-hand limits match the value at \(3\):
\[ \lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)\;\;\Longrightarrow\;\; k^{3}+3=\frac{16}{k^{2}-3}. \] Multiply out and simplify: \[ (k^{3}+3)(k^{2}-3)=16 \;\;\Longrightarrow\;\; k^{5}-3k^{3}+3k^{2}-25=0. \] For \(k>0\), the unique positive root of this equation is \[ k \approx 2.081. \] Therefore \(f\) is continuous when \(k\approx 2.081\).
✅ Answer: (A)