Practice Problem Solutions for 1/28
- Let \(h(x)=(x-5)^4+11\). Define functions \(f\) and \(g\) such that \(h=f\circ g\). (You can even try to come up with two different pairs of functions whose composition is \(h\).)
Let \(g(x)=x-5\) and \(f(x)=x^4+11\). Then,
\[
f\circ g(x)=f(x-5)=(x-5)^4+11=h(x).
\]
Note that both \(f\) and \(g\) have domain and range \(\mathbb{R}\), so the composition \(f\circ g\) is indeed defined.
We could also just consider \(f\) a shift of \(g\) by letting \(g(x)=(x-5)^4\) and \(f(x)=x+11\). In this case,
\[
f\circ g(x)=f((x-5)^4)=(x-5)^4+11.
\]
- Suppose a population of bacteria grows exponentially. Let \(s(t)\) be the number of cells in the colony after \(t\) days.
- What is an appropriate domain for the function \(s\)?
Since we are interested in time, should require \(t\geq 0\). That is, the domain of \(s\) is \([0,\infty) \).
- Suppose there are initially 50 cells and the number of cells triples each day. Find a formula for \(s(t)\), then use your formula to determine how many cells are in the colony after 10 days.
From the given information we know \(s\) is exponential so it looks something like \(s(t)=c\cdot a^t\). We need to figure out what the base \(a\) should be as well as what the constant \(c\) should be.
- We are given \(s(0)=50\). From our above formula for \(s\) this means \(50=c\cdot a^0=c\), since \(a^0=1\) for all \(a\). Thus, \(s(t)=50a^t\).
- We are also given that the number of cells triples each day. In particular this means \(s(t+1)=3s(t)\), for every \(t\). Plugging in the formula for \(s\) gives
\[
50a^{t+1}=3(50a^t).
\]
Simplifying the expression by dividing both sides by the common factor \(50a^t\) gives \(a=3\). Now we know \(c=50\) and \(a=3\), so \(s(t)=50\cdot 3^t\).
- Since we have a formula for \(s(t)\) we may project the size of the colony at any time in the future. In particular, after 10 days there will be \(s(10)=50\cdot3^{10}=2,952,450\) cells in the colony.