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Question 1. Find the local linear approximation of \(f(x)=2\sqrt{x}\) at \(x=4\).
Step 1: Use the local linear approximation formula \(L(x)=f(a)+f'(a)(x-a)\). Step 2: Here \(f(x)=2\sqrt{x}\) and \(a=4\). Step 3: Evaluate the function value: \(f(4)=2\sqrt{4}=4\). Step 4: Find the derivative: \(f'(x)=\frac{1}{\sqrt{x}}\). Step 5: Evaluate the derivative at \(x=4\): \(f'(4)=\frac{1}{\sqrt{4}}=\frac{1}{2}\). Step 6: Substitute into the formula: \(L(x)=4+\frac{1}{2}(x-4)\). Step 7: Simplify: \(L(x)=4+\frac{x}{2}-2=\frac{x}{2}+2\). Answer: C. \(y=\frac{1}{2}x+2\)
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Question 2. Find the local linear approximation of \(f(x)=2x^3\) at \(x=-2\).
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 the function value: \(f(-2)=2(-2)^3=-16\). Step 4: Find the derivative: \(f'(x)=6x^2\). Step 5: Evaluate the derivative at \(x=-2\): \(f'(-2)=6(-2)^2=24\). Step 6: Substitute into the formula: \(L(x)=-16+24(x-(-2))\). Step 7: Simplify: \(L(x)=-16+24(x+2)=24x+32\). Answer: A. \(y=24x+32\)
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Question 3. Approximating the integral \(\int_{3}^{5}x^3\,dx\) by using the Trapezoidal rule and using \(n=4\) intervals, yields a value for the integral of:
Step 1: Use the Trapezoidal rule: \(\int_a^b f(x)\,dx\approx \frac{\Delta x}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n-1})+f(x_n)\right]\). Step 2: Here \(a=3\), \(b=5\), and \(n=4\), so \(\Delta x=\frac{5-3}{4}=\frac{1}{2}\). Step 3: The points are \(x_0=3\), \(x_1=3.5\), \(x_2=4\), \(x_3=4.5\), and \(x_4=5\). Step 4: Since \(f(x)=x^3\), compute \(f(3)=27\), \(f(3.5)=\frac{343}{8}\), \(f(4)=64\), \(f(4.5)=\frac{729}{8}\), and \(f(5)=125\). Step 5: Substitute into the Trapezoidal rule: \(\int_3^5 x^3\,dx\approx \frac{1}{4}\left[27+2\left(\frac{343}{8}\right)+2(64)+2\left(\frac{729}{8}\right)+125\right]\). Step 6: Simplify: \(\frac{1}{4}\left[27+\frac{343}{4}+128+\frac{729}{4}+125\right]=137\). Answer: B. 137
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Question 4. Approximating the integral \(\int_{0}^{1}\tan x\,dx\) by using Simpson's rule and using \(n=4\) intervals, yields a value for the integral of:
Step 1: Use Simpson's rule: \(\int_a^b f(x)\,dx\approx \frac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)\right]\). Step 2: Here \(a=0\), \(b=1\), and \(n=4\), so \(\Delta x=\frac{1-0}{4}=\frac{1}{4}\). Step 3: The points are \(x_0=0\), \(x_1=\frac{1}{4}\), \(x_2=\frac{1}{2}\), \(x_3=\frac{3}{4}\), and \(x_4=1\). Step 4: Substitute \(f(x)=\tan x\) into Simpson's rule: \(\int_0^1 \tan x\,dx\approx \frac{1}{12}\left[\tan(0)+4\tan\left(\frac{1}{4}\right)+2\tan\left(\frac{1}{2}\right)+4\tan\left(\frac{3}{4}\right)+\tan(1)\right]\). Step 5: Evaluate the expression: \(\frac{1}{12}\left[0+4\tan\left(\frac{1}{4}\right)+2\tan\left(\frac{1}{2}\right)+4\tan\left(\frac{3}{4}\right)+\tan(1)\right]\approx 0.616\). Answer: C. 0.616
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