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The function 
f is defined by 
f(x)=x^(3)-5+cos(2x). 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)=x35+cos(2x) f(x)=x^{3}-5+\cos (2 x) . 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:

Full solution

Q. The function f f is defined by f(x)=x35+cos(2x) f(x)=x^{3}-5+\cos (2 x) . 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 and Slope: To find the equation of the tangent line at x=1x = 1, we need to calculate the derivative of f(x)f(x) to find the slope of the tangent line at that point.\newlineThe derivative of f(x)f(x) is f(x)=3x2sin(2x)2f'(x) = 3x^2 - \sin(2x) \cdot 2, using the power rule for x3x^3, the constant rule for 5-5, and the chain rule for cos(2x)\cos(2x).\newlineNow we need to evaluate f(x)f'(x) at x=1x = 1 to find the slope of the tangent line.
  2. Evaluate f(1)f'(1) for Slope: Calculating f(1)f'(1) gives us the slope of the tangent line at x=1x = 1.
    f(1)=3(1)2sin(21)2=3sin(2)2f'(1) = 3(1)^2 - \sin(2 \cdot 1) \cdot 2 = 3 - \sin(2) \cdot 2.
    Using a calculator, we find that sin(2)\sin(2) is approximately 0.9090.909, so f(1)30.9092f'(1) \approx 3 - 0.909 \cdot 2.
  3. Find y-coordinate at x=11: Now we perform the calculation for f(1)f'(1). f(1)30.909×2=31.818=1.182f'(1) \approx 3 - 0.909 \times 2 = 3 - 1.818 = 1.182 (rounded to three decimal places). So, the slope of the tangent line at x=1x = 1 is approximately 1.1821.182.
  4. Use Point-Slope Form: Next, we need to find the yy-coordinate of the point on the graph of f(x)f(x) where x=1x = 1. This is done by evaluating f(1)f(1).\newlinef(1)=135+cos(2×1)=15+cos(2)f(1) = 1^3 - 5 + \cos(2 \times 1) = 1 - 5 + \cos(2).\newlineUsing a calculator, we find that cos(2)\cos(2) is approximately 0.5400.540, so f(1)15+0.540f(1) \approx 1 - 5 + 0.540.
  5. Simplify Tangent Line Equation: Now we perform the calculation for f(1)f(1).f(1)15+0.540=4+0.540=3.460f(1) \approx 1 - 5 + 0.540 = -4 + 0.540 = -3.460 (rounded to three decimal places). So, the yy-coordinate of the point on the graph of f(x)f(x) at x=1x = 1 is approximately 3.460-3.460.
  6. Find y-intercept: With the slope of the tangent line and the point (1,3.460)(1, -3.460), we can use the point-slope form of the equation of a line to find the equation of the tangent line.\newlineThe point-slope form is yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line.\newlineSubstituting the values we have, the equation becomes y(3.460)=1.182(x1)y - (-3.460) = 1.182(x - 1).
  7. Find y-intercept: With the slope of the tangent line and the point (1,3.460)(1, -3.460), we can use the point-slope form of the equation of a line to find the equation of the tangent line.\newlineThe point-slope form is yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line.\newlineSubstituting the values we have, the equation becomes y(3.460)=1.182(x1)y - (-3.460) = 1.182(x - 1). Simplifying the equation of the tangent line, we get:\newliney+3.460=1.182x1.182y + 3.460 = 1.182x - 1.182.\newlineNow we subtract 3.4603.460 from both sides to solve for yy.\newliney=1.182x1.1823.460y = 1.182x - 1.182 - 3.460.
  8. Find y-intercept: With the slope of the tangent line and the point (1,3.460)(1, -3.460), we can use the point-slope form of the equation of a line to find the equation of the tangent line.\newlineThe point-slope form is yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line.\newlineSubstituting the values we have, the equation becomes y(3.460)=1.182(x1)y - (-3.460) = 1.182(x - 1). Simplifying the equation of the tangent line, we get:\newliney+3.460=1.182x1.182y + 3.460 = 1.182x - 1.182.\newlineNow we subtract 3.4603.460 from both sides to solve for yy.\newliney=1.182x1.1823.460y = 1.182x - 1.182 - 3.460. Finally, we perform the subtraction to find the y-intercept of the tangent line.\newliney=1.182x4.642y = 1.182x - 4.642 (rounded to three decimal places).\newlineThis is the equation of the tangent line at yy1=m(xx1)y - y_1 = m(x - x_1)00 for the function yy1=m(xx1)y - y_1 = m(x - x_1)11.

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