Question

1. let $f(x)=ln(x - 3)$. which is greater: the instantaneous rate of change of $f$ at $x = 4$ or the instantaneous rate of change of $f$ at $x = 10$? use the graph of $f$ to justify your answer.
2. the weight of a baby duck, $w(t)$, $t$ days after its birth can be modeled by the function $w(t)=100 - 80e^{-0.2t}$, where $w(t)$ is in grams. find the average rate of change in the baby ducks weight over its first month of life (assume the month has 31 days). round to the nearest thousandth.

Explanation:

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Step1: Find the derivative of $f(x)$

The derivative of $y = \ln(u)$ is $y^\prime=\frac{u^\prime}{u}$. For $f(x)=\ln(x - 3)$, let $u=x - 3$, then $u^\prime = 1$. So $f^\prime(x)=\frac{1}{x - 3}$.

Step2: Calculate the instantaneous - rate of change at $x = 4$

Substitute $x = 4$ into $f^\prime(x)$: $f^\prime(4)=\frac{1}{4 - 3}=1$.

Step3: Calculate the instantaneous - rate of change at $x = 10$

Substitute $x = 10$ into $f^\prime(x)$: $f^\prime(10)=\frac{1}{10 - 3}=\frac{1}{7}\approx0.143$.

Step4: Compare the two values

Since $1>\frac{1}{7}$, the instantaneous rate of change of $f$ at $x = 4$ is greater.
Justification using the graph: The derivative of a function represents the slope of the tangent line to the graph of the function. For $y = \ln(x - 3)$, the graph is a logarithmic function shifted 3 units to the right. The function is concave - down, and the slopes of the tangent lines decrease as $x$ increases. So the slope of the tangent line at $x = 4$ is steeper than the slope of the tangent line at $x = 10$.

Step1: Recall the formula for the average rate of change

The average rate of change of a function $y = W(t)$ over the interval $[a,b]$ is $\frac{W(b)-W(a)}{b - a}$. Here, $a = 0$, $b = 31$, and $W(t)=100 - 80e^{-0.2t}$.

Step2: Calculate $W(0)$

Substitute $t = 0$ into $W(t)$: $W(0)=100-80e^{-0.2\times0}=100 - 80=20$.

Step3: Calculate $W(31)$

$W(31)=100-80e^{-0.2\times31}=100-80e^{-6.2}$. Since $e^{-6.2}\approx0.00202$, then $W(31)=100-80\times0.00202=100 - 0.1616 = 99.8384$.

Step4: Calculate the average rate of change

$\frac{W(31)-W(0)}{31 - 0}=\frac{99.8384 - 20}{31}=\frac{79.8384}{31}\approx2.575$.

Answer:

The instantaneous rate of change of $f$ at $x = 4$ is greater.

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