EBUS CAL12 Chapter 7 Quiz 2-4 (lesson 3,4)

Step 1. Use the local linear approximation formula \(L(x)=f(a)+f'(a)(x-a)\). Step 2. Here \(f(x)=2x^3\) and \(a=-2\). Step 3. Evaluate \(f(-2)=2(-2)^3=-16\). Step 4. Differentiate: \(f'(x)=6x^2\), so \(f'(-2)=6(-2)^2=24\). Step 5. Substitute into the formula: \(L(x)=-16+24(x-(-2))\). Step 6. Simplify: \(L(x)=-16+24(x+2)=24x+32\). Answer: D

Step 1. Use \(L(x)=f(a)+f'(a)(x-a)\). Step 2. Here \(f(x)=\frac{1}{x}\) and \(a=1\). Step 3. Evaluate \(f(1)=1\). Step 4. Differentiate: \(f'(x)=-\frac{1}{x^2}\), so \(f'(1)=-1\). Step 5. Substitute into the formula: \(L(x)=1-1(x-1)\). Step 6. Simplify: \(L(x)=1-x+1=-x+2\). Answer: A

Step 1. Use the Trapezoidal Rule: \(\int_a^b f(x)\,dx\approx \frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)]\). Step 2. Here \(a=2\), \(b=5\), and \(n=4\), so \(\Delta x=\frac{5-2}{4}=\frac{3}{4}\). Step 3. The points are \(x_0=2\), \(x_1=\frac{11}{4}\), \(x_2=\frac{7}{2}\), \(x_3=\frac{17}{4}\), and \(x_4=5\). Step 4. Since \(f(x)=\sqrt{x-1}\), substitute the function values into the trapezoidal rule. Step 5. The approximation becomes \(\frac{3}{8}[1+\sqrt{7}+\sqrt{10}+\sqrt{13}+2]\). Step 6. Evaluate: \(\frac{3}{8}[1+\sqrt{7}+\sqrt{10}+\sqrt{13}+2]\approx 4.655\). Answer: C

Step 1. Use Simpson's Rule: \(\int_a^b f(x)\,dx\approx \frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)]\). Step 2. Here \(a=0\), \(b=4\), and \(n=4\), so \(\Delta x=\frac{4-0}{4}=1\). Step 3. The points are \(x_0=0\), \(x_1=1\), \(x_2=2\), \(x_3=3\), and \(x_4=4\). Step 4. With \(f(x)=5x^2+1\), the function values are \(f(0)=1\), \(f(1)=6\), \(f(2)=21\), \(f(3)=46\), and \(f(4)=81\). Step 5. Substitute into Simpson's Rule: \(\frac{1}{3}[1+4(6)+2(21)+4(46)+81]\). Step 6. Simplify: \(\frac{1}{3}[1+24+42+184+81]=\frac{332}{3}\approx 110.667\). Answer: C