Exercise 1 · Functions and derivatives — Social media followers

Reading values, monotonicity intervals, local extrema and sketching a cubic on a closed interval.

Maximum score · 2.5 points

A doctor starts doing health outreach on social media. As he publishes content on this topic, he sees his number of followers grow. At a certain moment he records a controversial video and loses a small number of followers, but a few days later he publishes another one that is very successful and, from then on, his follower count grows again. The function describing his number of followers as a function of time $t$, measured in weeks, is $f(t) = 10t^{3} - 120t^{2} + 450t + 700$, with $t \in [0, 10]$.

How many followers does he have at the start? How many after 10 weeks? Study the intervals where the function increases and decreases and find its local extrema. Sketch the graph using the information obtained. 1.5 p
In which week does he post the controversial video, and how many followers does he have at that moment? In which week does he post the very successful video? Over the 10 weeks, when does he have the most followers? 1 p

Bachillerato CCSS · Block D — Algebraic sense Derivatives Monotonicity and extrema Curve sketching

Step-by-step solution

Key idea: the monotonicity of $f$ is given by the sign of $f'$; local extrema are the zeros of $f'$ where the sign changes. On a closed interval, the absolute maximum can be a local extremum or an endpoint: values must be compared.

a) Initial and final values, monotonicity and extrema

Followers at the start and after 10 weeks:

f(0) = 700, \qquad f(10) = 10(1000) - 120(100) + 450(10) + 700 = 10000 - 12000 + 4500 + 700 = 3200.

Differentiate and factor:

f'(t) = 30t^{2} - 240t + 450 = 30\,(t^{2} - 8t + 15) = 30\,(t-3)(t-5).

The zeros of $f'$ are $t = 3$ and $t = 5$. We study the sign of $f'$ on $[0, 10]$:

f' > 0 \text{ on } [0, 3), \qquad f' < 0 \text{ on } (3, 5), \qquad f' > 0 \text{ on } (5, 10].

Hence $f$ is increasing on $[0,3) \cup (5,10]$ and decreasing on $(3,5)$.

Local extrema: at $t=3$ there is a local maximum with $f(3) = 270 - 1080 + 1350 + 700 = 1240$; at $t=5$ a local minimum with $f(5) = 1250 - 3000 + 2250 + 700 = 1200$.

Sketch through the key points $(0,700)$, $(3,1240)$, $(5,1200)$ and $(10,3200)$:

$f(0)=700$ and $f(10)=3200$ followers. Increasing on $[0,3)\cup(5,10]$, decreasing on $(3,5)$; local maximum $(3, 1240)$ and local minimum $(5, 1200)$.

b) Interpreting the extrema

The controversial video is posted when he starts losing followers, i.e. at the local maximum: week 3, with $f(3) = 1240$ followers.

The very successful video is posted when the follower count starts growing again, i.e. at the local minimum: week 5.

To find the moment with the most followers we compare the local maximum with the endpoints:

f(3) = 1240 \quad \text{vs.} \quad f(10) = 3200.

Controversial video: week 3, with 1240 followers. Successful video: week 5. The follower count peaks at the end, in week 10, with 3200 followers.