Question

question 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. answer the instantaneous rate of change of $f(x)$ at $x = 4$ is greater because the graph of $f$ is always increasing but getting flatter, so the tangent line at $x = 4$ is steeper than the tangent line at $x = 10$. the graph of $y = h(x)$ is shown. find the average rate of change of $h$ on the interval $-4,3$.

Explanation:

Step1: Recall the derivative formula

The derivative of $f(x)=\ln(x - 3)$ is $f'(x)=\frac{1}{x - 3}$ by the chain - rule (since the derivative of $\ln(u)$ is $\frac{1}{u}$ and $u=x - 3$ with $u'=1$).

Step2: Calculate the derivative at $x = 4$

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

Step3: Calculate the derivative at $x = 10$

Substitute $x = 10$ into $f'(x)$: $f'(10)=\frac{1}{10 - 3}=\frac{1}{7}$.

Step4: Compare the values

Since $1>\frac{1}{7}$, the instantaneous rate of change of $f$ at $x = 4$ is greater.

Answer:

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