The number of **Facebook** members from the start of 2006 to mid-2008 can be approximated by

*S*(*t*) = 15*t*2 − 76*t* + 101 million members  (2 ≤ *t* ≤ 4.5)

in year *t* (*t* = 0 represents the start of 2004).†

A coordinate plane is labeled "Facebook membership (millions)." There is a curve and five points.

The curve enters the window in the first quadrant, goes down and right becoming less steep, changes direction at the approximate point (2.53, 4.7), goes up and right becoming more steep, and exits the window in the first quadrant.

The pattern of points goes up and right becoming more steep from (2, 6) to (4.5, 59).

(a)

Compute *S'*(*t*).

*S'*(*t*) = 
 
According to the model, how fast was **Facebook** membership changing at the start of 2007 (*t* = 3)? (Be careful to give correct units of measurement.)

*S'*(3) =  million members per year

(b)

According to the model, **Facebook** memberships did which of the following?

decreased at a faster and faster rate during 2007 (the interval [3, 4])increased at a slower and slower rate during 2007 (the interval [3, 4])    decreased at a slower and slower rate during 2007 (the interval [3, 4])increased at a faster and faster rate during 2007 (the interval [3, 4])




Let's go through each part step by step:

### (a) Compute \( S'(t) \)

The function for Facebook membership is given as:

\[
S(t) = 15t^2 - 76t + 101
\]

To find \( S'(t) \), we need to differentiate \( S(t) \) with respect to \( t \):

\[
S'(t) = \frac{d}{dt} \left( 15t^2 - 76t + 101 \right)
\]

Apply the power rule to each term:

- The derivative of \( 15t^2 \) is \( 30t \).
- The derivative of \( -76t \) is \( -76 \).
- The derivative of the constant \( 101 \) is \( 0 \).

So, the derivative is:

\[
S'(t) = 30t - 76
\]

### Now, calculate \( S'(3) \):

\[
S'(3) = 30(3) - 76 = 90 - 76 = 14
\]

Thus, \( S'(3) = 14 \) million members per year. This means that at the start of 2007 (\( t = 3 \)), the number of Facebook members was increasing at a rate of 14 million members per year.

### (b) Behavior of Facebook Memberships during 2007 (the interval [3, 4])

To understand how the Facebook membership changes during 2007, we need to analyze the behavior of \( S'(t) \) on the interval [3, 4].

The derivative is:

\[
S'(t) = 30t - 76
\]

This is a linear function with a positive slope (30), meaning that as \( t \) increases, \( S'(t) \) increases. Specifically:

- At \( t = 3 \), \( S'(3) = 14 \), which we already calculated.
- At \( t = 4 \), \( S'(4) = 30(4) - 76 = 120 - 76 = 44 \).

Since \( S'(t) \) is increasing from 14 to 44 over the interval [3, 4], this means that the Facebook membership is increasing at a faster and faster rate during 2007.

Thus, the correct answer is:

**Increased at a faster and faster rate during 2007 (the interval [3, 4]).**

---

Would you like to explore further details or have any questions? Here are some related questions:

1. What would the membership growth rate be at mid-2008 (\( t = 4.5 \))?
2. Can we find the time when the membership growth rate was zero?
3. How would the membership change if the coefficients of \( t \) were different?
4. What does the concavity of the curve tell us about membership growth over time?
5. What is the total change in membership from the start of 2007 to the end of 2007?

**Tip:** Analyzing the second derivative \( S''(t) \) helps understand whether a function is speeding up or slowing down in its growth or decline.

The number of Facebook members from the start of 2006 to mid-2008 can be approximated by S(t) = 15t^2 − 76t + 101 million members (2 ≤ t ≤ 4.5) in year t (t = 0 represents the start of 2004). Compute S'(t) and analyze Facebook membership growth during 2007 (the interval [3, 4]).

Explore the model approximating Facebook membership growth from 2006 to mid-2008, using differentiation. Learn how to calculate the rate of change and analyze Facebook's membership behavior during 2007. Ideal for students in Grades 11-12.

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