Question

1. calculator allowed

x	2	5	7	8
f(x)	10	30	40	20

the function ( f ) is continuous on the closed interval (2,8) and has values that are given in the table above. using the subintervals (2,5), (5,7), and (7,8), what is the trapezoidal approximation of ( int_{2}^{8} f(x) dx )?
(a) 110
(b) 130
(c) 160
(d) 190
(e) 210

calculator allowed
which of the following is an equation of the line tangent to the graph of ( f(x) = x^4 + 2x^2 ) at the point where ( f(x) = 1 )?
(a) ( y = 8x - 5 )
(b) ( y = x + 7 )
(c) ( y = x + 0.763 )
(d) ( y = x - 0.122 )
(e) ( y = x - 2.146 )

Explanation:

First Sub - Question (Trapezoidal Approximation)

Step 1: Recall the trapezoidal rule formula

The trapezoidal rule for approximating $\int_{a}^{b}f(x)dx$ with subintervals $[x_0,x_1],[x_1,x_2],\cdots,[x_{n - 1},x_n]$ is $\int_{a}^{b}f(x)dx\approx\frac{\Delta x_1}{2}(f(x_0)+f(x_1))+\frac{\Delta x_2}{2}(f(x_1)+f(x_2))+\cdots+\frac{\Delta x_n}{2}(f(x_{n - 1})+f(x_n))$, where $\Delta x_i=x_i - x_{i - 1}$.

Step 2: Determine the subintervals and their widths

We have three subintervals: $[2,5]$, $[5,7]$, and $[7,8]$.

• For $[2,5]$, $\Delta x_1 = 5 - 2=3$, $f(2) = 10$, $f(5)=30$.
• For $[5,7]$, $\Delta x_2=7 - 5 = 2$, $f(5)=30$, $f(7)=40$.
• For $[7,8]$, $\Delta x_3=8 - 7=1$, $f(7)=40$, $f(8)=20$.

Step 3: Apply the trapezoidal rule

\[

$$\begin{align*} \int_{2}^{8}f(x)dx&\approx\frac{3}{2}(10 + 30)+\frac{2}{2}(30 + 40)+\frac{1}{2}(40+20)\\ &=\frac{3}{2}(40)+1\times(70)+\frac{1}{2}(60)\\ &=60 + 70+30\\ &=160 \end{align*}$$

\]

Step 1: Find the derivative of $f(x)$

Given $f(x)=x^{4}+2x^{2}$, then $f^{\prime}(x)=4x^{3}+4x$.

Step 2: Solve $f^{\prime}(x) = 1$

We need to solve the equation $4x^{3}+4x=1$, or $4x^{3}+4x - 1=0$. Using a calculator (since calculator is allowed), we can find the real root of this cubic equation. Let's call the root $x = a$. By using numerical methods (e.g., Newton - Raphson method) or a calculator, we find that $x\approx0.222$.

Step 3: Find the value of $f(x)$ at $x = a$

$f(a)=a^{4}+2a^{2}$. Substituting $a\approx0.222$, we get $f(0.222)\approx(0.222)^{4}+2\times(0.222)^{2}\approx0.0023 + 2\times0.0493\approx0.0023+0.0986 = 0.1009$.

Step 4: Find the equation of the tangent line

The equation of a tangent line at $x = a$ is $y - f(a)=f^{\prime}(a)(x - a)$. Since $f^{\prime}(a) = 1$, the equation becomes $y - f(a)=1\times(x - a)$, or $y=x - a+f(a)$. Substituting $a\approx0.222$ and $f(a)\approx0.1009$, we get $y=x-0.222 + 0.1009=x - 0.1211\approx x-0.122$.

Answer:

C

Second Sub - Question (Tangent Line)