Online Video Views The number of views, in millions, of a viral video days after posting is modeled by According to this model, on which day is the instantaneous rate at which the number of views is increasing the greatest?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 147 | Free SAT Prep | Aniko

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Question 147·Hard·Nonlinear Functions

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Online Video Views

The number of views, in millions, of a viral video $t$ days after posting is modeled by

V(t) = \frac{1.2}{1 + 4e^{-0.3t}}.

According to this model, on which day is the instantaneous rate at which the number of views is increasing the greatest?

When you see a logistic function of the form $\dfrac{L}{1+Ae^{-kt}}$ and are asked when the quantity is increasing fastest, recognize that this is the point where the graph is steepest—its inflection point—which occurs when the function value is half of its limiting value $L$ . On the SAT, avoid full calculus: instead, (1) identify the limiting value from the formula, (2) set the function equal to half of that value and solve for $t$ using algebra and logarithms, and (3) match the resulting time to the closest answer choice; you can quickly confirm with your graphing calculator by checking where the curve looks steepest.

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Think about the shape of the graph

Look at the form $V(t) = \dfrac{1.2}{1 + 4e^{-0.3t}}$ . As $t$ gets very large, where does $V(t)$ seem to be heading? What does that tell you about the overall shape of the graph?

Where is a logistic curve steepest?

For an S-shaped (logistic) curve that starts near 0 and levels off near a maximum, does it grow fastest at the very beginning, in the middle, or near the end? Think about where the graph looks steepest.

Use the maximum value

The function levels off near $1.2$ million views. For logistic growth, the point where the growth rate is largest occurs when the function value is halfway to that maximum. What value of $V(t)$ is halfway to $1.2$ , and how can you solve $V(t)$ equal to that value for $t$ ?

Connect your $t$ value to a day

After solving for $t$ , compare your result to the whole-number days in the answer choices. Which day is it closest to?

Desmos Guide

Graph the function

In Desmos, enter the function as

y = \frac{1.2}{1 + 4e^{-0.3x}}.

Make sure $x$ represents days and $y$ represents views in millions.

Adjust the viewing window

Set the $x$ -axis to show from about $x=0$ to $x=20$ and the $y$ -axis from $y=0$ to about $y=1.3$ . This will let you clearly see the S-shaped curve as it rises and levels off.

Locate where the graph is steepest

Visually inspect the graph and find the $x$ -value where the curve is steepest (where it is rising the fastest). Look along the $x$ -axis to see which whole-number day (among 2, 5, 8, 14) this steepest point is closest to.

Optional: Use a table to compare growth near each choice

Add a table for the function and plug in values of $x$ near each answer choice (for example, around 2, 5, 8, and 14). Compare how much $y$ increases per day in each region; the day whose nearby $x$ -values produce the largest increases corresponds to the greatest instantaneous rate of increase.

Step-by-step Explanation

Understand what “instantaneous rate of increase” means

The instantaneous rate at which the views are increasing is the slope of the curve $V(t)$ at a specific time $t$ (in calculus terms, this is $V'(t)$ ). The question is asking: for which day is the slope of the graph of $V(t)$ the greatest?

Recognize the type and shape of the function

The function

V(t) = \frac{1.2}{1 + 4e^{-0.3t}}

is a logistic function. Logistic curves:

Start near 0,
Increase in an S-shape,
Level off near a maximum value (here, $1.2$ million views).

For a logistic curve, the rate of increase is:

Small at the beginning,
Largest at the middle of the S-curve,
Then gets smaller again as it levels off.

This steepest point (maximum instantaneous rate) occurs when the function is half of its maximum value.

Find when the video is at half of its maximum views

The maximum (long-term) value of $V(t)$ is $1.2$ million views, so half of that is

\frac{1.2}{2} = 0.6 \text{ million views}.

Set $V(t)$ equal to $0.6$ and solve for $t$ :

0.6 = \frac{1.2}{1 + 4e^{-0.3t}}.

Multiply both sides by $1 + 4e^{-0.3t}$ :

0.6\bigl(1 + 4e^{-0.3t}\bigr) = 1.2.

Distribute $0.6$ :

0.6 + 2.4e^{-0.3t} = 1.2.

Subtract $0.6$ from both sides:

2.4e^{-0.3t} = 0.6.

Divide both sides by $2.4$ :

e^{-0.3t} = \frac{0.6}{2.4} = \frac{1}{4}.

Take the natural log of both sides:

-0.3t = \ln\left(\frac{1}{4}\right) = -\ln 4,

t = \frac{\ln 4}{0.3} \approx \frac{1.386}{0.3} \approx 4.6.

Match the time to the closest answer choice

The time when the instantaneous rate of increase is greatest is about $t\approx 4.6$ days after posting. That is closest to Day 5, so the correct answer is Day 5.

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Desmos Guide

Step-by-step Explanation

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