6. let $f(x)=ln(x - 3)$. which is greater: the instantaneous rate of change of $f$ at $x = 4$ or the…
6. 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.\n7. 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.
Answer
6.
Explanation:
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 $x = 10$.
Answer: The instantaneous rate of change of $f$ at $x = 4$ is greater.
7.
Explanation:
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
$\frac{W(31)-W(0)}{31 - 0}=\frac{99.8384 - 20}{31}=\frac{79.8384}{31}\approx2.575$.