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The function f is defined by f(x)=x^(3)+3cos(2x-4). Use a calculator to write the equation of the line tangent to the graph of f when x=-1. You should round all decimals to 3 places. Answer:

The function f f is defined by f(x)=x3+3cos⁡(2x−4) f(x)=x^{3}+3 \cos (2 x-4) . Use a calculator to write the equation of the line tangent to the graph of f f when x=−1 x=-1 . You should round all decimals to 33 places.\newlineAnswer:

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Q. The function f f is defined by f(x)=x3+3cos⁡(2x−4) f(x)=x^{3}+3 \cos (2 x-4) . Use a calculator to write the equation of the line tangent to the graph of f f when x=−1 x=-1 . You should round all decimals to 33 places.\newlineAnswer:
  1. Calculate Derivative: To find the equation of the tangent line at x=−1x = -1, we first need to calculate the derivative of f(x)f(x), which will give us the slope of the tangent line at any point xx. The derivative of f(x)=x3+3cos⁡(2x−4)f(x) = x^3 + 3\cos(2x - 4) is f′(x)=3x2−6sin⁡(2x−4)f'(x) = 3x^2 - 6\sin(2x - 4).
  2. Find Slope at x=−1x = -1: Now we need to evaluate the derivative at x=−1x = -1 to find the slope of the tangent line at that point.f′(−1)=3(−1)2−6sin⁡(2(−1)−4)=3−6sin⁡(−6)f'(-1) = 3(-1)^2 - 6\sin(2(-1) - 4) = 3 - 6\sin(-6).Using a calculator, we find that sin⁡(−6)≈−0.2794\sin(-6) \approx -0.2794 (rounded to four decimal places for intermediate calculation).
  3. Find y-coordinate at x=−1x = -1: Now we plug the value of sin⁡(−6)\sin(-6) into the derivative to get the slope at x=−1x = -1.f′(−1)=3−6(−0.2794)≈3+1.6764=4.6764f'(-1) = 3 - 6(-0.2794) \approx 3 + 1.6764 = 4.6764 (rounded to four decimal places for intermediate calculation). We round the slope to three decimal places as instructed: slope ≈4.676\approx 4.676.
  4. Use Point-Slope Form: Next, we need to find the y-coordinate of the point on the graph of f(x)f(x) at x=−1x = -1 to use it in the point-slope form of the equation of the tangent line.f(−1)=(−1)3+3cos⁡(2(−1)−4)=−1+3cos⁡(−6)f(-1) = (-1)^3 + 3\cos(2(-1) - 4) = -1 + 3\cos(-6).Using a calculator, we find that cos⁡(−6)≈0.9602\cos(-6) \approx 0.9602 (rounded to four decimal places for intermediate calculation).
  5. Convert to Slope-Intercept Form: Now we plug the value of cos⁡(−6)\cos(-6) into the function to get the yy-coordinate at x=−1x = -1.f(−1)=−1+3(0.9602)≈−1+2.8806=1.8806f(-1) = -1 + 3(0.9602) \approx -1 + 2.8806 = 1.8806 (rounded to four decimal places for intermediate calculation). We round the yy-coordinate to three decimal places as instructed: yy-coordinate ≈1.881\approx 1.881.
  6. Convert to Slope-Intercept Form: Now we plug the value of cos⁡(−6)\cos(-6) into the function to get the y-coordinate at x=−1x = -1.f(−1)=−1+3(0.9602)≈−1+2.8806=1.8806f(-1) = -1 + 3(0.9602) \approx -1 + 2.8806 = 1.8806 (rounded to four decimal places for intermediate calculation). We round the y-coordinate to three decimal places as instructed: y-coordinate ≈1.881\approx 1.881.We now have the slope of the tangent line (m≈4.676m \approx 4.676) and a point on the tangent line (x=−1x = -1, y≈1.881y \approx 1.881). We can use the point-slope form of the equation of a line to write the equation of the tangent line: y−y1=m(x−x1)y - y_1 = m(x - x_1). Substituting the known values, we get: y−1.881=4.676(x−(−1))y - 1.881 = 4.676(x - (-1)).
  7. Convert to Slope-Intercept Form: Now we plug the value of cos⁡(−6)\cos(-6) into the function to get the y-coordinate at x=−1x = -1.f(−1)=−1+3(0.9602)≈−1+2.8806=1.8806f(-1) = -1 + 3(0.9602) \approx -1 + 2.8806 = 1.8806 (rounded to four decimal places for intermediate calculation). We round the y-coordinate to three decimal places as instructed: y-coordinate ≈1.881\approx 1.881.We now have the slope of the tangent line (m≈4.676m \approx 4.676) and a point on the tangent line (x=−1x = -1, y≈1.881y \approx 1.881). We can use the point-slope form of the equation of a line to write the equation of the tangent line: y−y1=m(x−x1)y - y_1 = m(x - x_1). Substituting the known values, we get: y−1.881=4.676(x−(−1))y - 1.881 = 4.676(x - (-1)).Simplify the equation to get it into the slope-intercept form y=mx+by = mx + b. x=−1x = -100. Now, add x=−1x = -111 to both sides to isolate x=−1x = -122. x=−1x = -133.
  8. Convert to Slope-Intercept Form: Now we plug the value of cos⁡(−6)\cos(-6) into the function to get the y-coordinate at x=−1x = -1.f(−1)=−1+3(0.9602)≈−1+2.8806=1.8806f(-1) = -1 + 3(0.9602) \approx -1 + 2.8806 = 1.8806 (rounded to four decimal places for intermediate calculation). We round the y-coordinate to three decimal places as instructed: y-coordinate ≈1.881\approx 1.881.We now have the slope of the tangent line (m≈4.676m \approx 4.676) and a point on the tangent line (x=−1x = -1, y≈1.881y \approx 1.881). We can use the point-slope form of the equation of a line to write the equation of the tangent line: y−y1=m(x−x1)y - y_1 = m(x - x_1). Substituting the known values, we get: y−1.881=4.676(x−(−1))y - 1.881 = 4.676(x - (-1)).Simplify the equation to get it into the slope-intercept form y=mx+by = mx + b. x=−1x = -100. Now, add x=−1x = -111 to both sides to isolate x=−1x = -122. x=−1x = -133.Combine like terms to get the final equation of the tangent line. x=−1x = -144. We round the constant term to three decimal places as instructed: x=−1x = -155.

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