How to estimate a derivative, use the Intermediate Value Theorem, and compute trapezoidal and accumulation approximations

Key concept: This is the kind of free-response problem where each part tests a different skill. The trick is to stay organized: local rate, existence, numerical integration, then a model-based accumulation calculation.

Step by step

A. Approximate $R'(1)$

Use the average rate of change of $R$ on $[0,2]$:

$R'(1)\approx \frac{R(2)-R(0)}{2-0}=\frac{100-90}{2}=5.$

So

$\boxed{R'(1)\approx 5\text{ words per minute}^2}$

B. Must there be a value $c$ such that $R(c)=155$?

Yes.

From the table:

$R(8)=150 \quad\text{and}\quad R(10)=162.$

Because $R$ is differentiable, it is continuous on $[8,10]$. Since 155 lies between 150 and 162, the Intermediate Value Theorem guarantees at least one value $c$ in $(8,10)$ with

$R(c)=155.$

So the answer is

$\boxed{\text{Yes, by the Intermediate Value Theorem.}}$

C. Trapezoidal sum for $\int_0^{10}R(t)\,dt$

Use the intervals $[0,2]$, $[2,8]$, and $[8,10]$.

$T=\frac{2}{2}[R(0)+R(2)] + \frac{6}{2}[R(2)+R(8)] + \frac{2}{2}[R(8)+R(10)]$

$=1(90+100)+3(100+150)+1(150+162)$

$=190+750+312=1252.$

So

$\boxed{\int_0^{10}R(t)\,dt\approx 1252\text{ words}}$

D. Words read by the teacher

The teacher's rate is

$W(t)=-\frac{3}{10}t^2+8t+100.$

The total words read in 10 minutes is

$\int_0^{10}W(t)\,dt.$

Compute:

=\left[-\frac{1}{10}t^3+4t^2+100t\right]_0^{10}.$$ At $t=10$: $$-\frac{1}{10}(1000)+4(100)+100(10)=-100+400+1000=1300.$$ At $t=0$: $$0.$$ So the teacher read $$\boxed{1300\text{ words}}.$$ ### Final answers - $\boxed{R'(1)\approx 5}$ - $\boxed{\text{Yes, by IVT}}$ - $\boxed{1252\text{ words}}$ - $\boxed{1300\text{ words}}$ ### Pitfall alert For part A, don't divide by 1 just because you are asked for $R'(1)$; the table only gives an estimate, so you use the slope over the nearest interval. For part C, another frequent mistake is forgetting to multiply each trapezoid average by its interval width. That changes the answer a lot. ### Try different conditions If the table had included more points, a Simpson's Rule estimate might be more accurate than the trapezoidal sum. If part D had asked for the average reading rate instead of total words, you would divide the integral by 10 after computing it. ### Further reading Intermediate Value Theorem, trapezoidal sum, accumulation function