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The function f is defined by f(x)=x^(2)-2sin(3x+1). Use a calculator to write the equation of the line tangent to the graph of f when x=-1.5. You should round all decimals to 3 places. Answer:

The function f f is defined by f(x)=x2−2sin⁡(3x+1) f(x)=x^{2}-2 \sin (3 x+1) . Use a calculator to write the equation of the line tangent to the graph of f f when x=−1.5 x=-1.5 . You should round all decimals to 33 places.\newlineAnswer:

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Q. The function f f is defined by f(x)=x2−2sin⁡(3x+1) f(x)=x^{2}-2 \sin (3 x+1) . Use a calculator to write the equation of the line tangent to the graph of f f when x=−1.5 x=-1.5 . You should round all decimals to 33 places.\newlineAnswer:
  1. Calculate Derivative: To find the equation of the tangent line at a specific point, we need to calculate the derivative of the function to find the slope of the tangent line at that point. The derivative of f(x)=x2−2sin⁡(3x+1)f(x) = x^2 - 2\sin(3x+1) is f′(x)=2x−2(3cos⁡(3x+1))f'(x) = 2x - 2(3\cos(3x+1)) by using the power rule for x2x^2 and the chain rule for −2sin⁡(3x+1)-2\sin(3x+1).
  2. Evaluate Derivative at x=−1.5x=-1.5: Now we need to evaluate the derivative at x=−1.5x = -1.5 to find the slope of the tangent line at that point. So we calculate f′(−1.5)=2(−1.5)−2(3cos⁡(3(−1.5)+1))f'(-1.5) = 2(-1.5) - 2(3\cos(3(-1.5)+1)).
  3. Find Slope and Point of Tangency: Using a calculator, we find that cos⁡(3(−1.5)+1)=cos⁡(−4.5+1)=cos⁡(−3.5)\cos(3(-1.5)+1) = \cos(-4.5+1) = \cos(-3.5). We round this to three decimal places.
  4. Calculate y-coordinate at x=−1-1.55: After rounding, let's assume cos⁡(−3.5)≈C\cos(-3.5) \approx C (where CC is the rounded value to three decimal places). Now we calculate f′(−1.5)=2(−1.5)−2(3C)=−3−6Cf'(-1.5) = 2(-1.5) - 2(3C) = -3 - 6C.
  5. Write Equation of Tangent Line: Next, we need to find the yy-coordinate of the function ff at x=−1.5x = -1.5 to determine the point of tangency. We calculate f(−1.5)=(−1.5)2−2sin⁡(3(−1.5)+1)f(-1.5) = (-1.5)^2 - 2\sin(3(-1.5)+1).
  6. Simplify Equation: Using a calculator, we find that sin⁡(3(−1.5)+1)=sin⁡(−4.5+1)=sin⁡(−3.5)\sin(3(-1.5)+1) = \sin(-4.5+1) = \sin(-3.5). We round this to three decimal places.
  7. Isolate yy in Slope-Intercept Form: After rounding, let's assume sin⁡(−3.5)≈S\sin(-3.5) \approx S (where SS is the rounded value to three decimal places). Now we calculate f(−1.5)=(−1.5)2−2S=2.25−2Sf(-1.5) = (-1.5)^2 - 2S = 2.25 - 2S.
  8. Combine Like Terms: With the slope of the tangent line −3−6C-3 - 6C and the point of tangency (−1.5,2.25−2S)(-1.5, 2.25 - 2S), we can use the point-slope form of the equation of a line to write the equation of the tangent line: y−(2.25−2S)=(−3−6C)(x−(−1.5))y - (2.25 - 2S) = (-3 - 6C)(x - (-1.5)).
  9. Combine Like Terms: With the slope of the tangent line −3−6C-3 - 6C and the point of tangency (−1.5,2.25−2S)(-1.5, 2.25 - 2S), we can use the point-slope form of the equation of a line to write the equation of the tangent line: y−(2.25−2S)=(−3−6C)(x−(−1.5))y - (2.25 - 2S) = (-3 - 6C)(x - (-1.5)). Simplifying the equation, we get y−2.25+2S=(−3−6C)(x+1.5)y - 2.25 + 2S = (-3 - 6C)(x + 1.5). Expanding the right side, we get y−2.25+2S=−3x−4.5−6Cx−9Cy - 2.25 + 2S = -3x - 4.5 - 6Cx - 9C.
  10. Combine Like Terms: With the slope of the tangent line −3−6C-3 - 6C and the point of tangency (−1.5,2.25−2S)(-1.5, 2.25 - 2S), we can use the point-slope form of the equation of a line to write the equation of the tangent line: y−(2.25−2S)=(−3−6C)(x−(−1.5))y - (2.25 - 2S) = (-3 - 6C)(x - (-1.5)). Simplifying the equation, we get y−2.25+2S=(−3−6C)(x+1.5)y - 2.25 + 2S = (-3 - 6C)(x + 1.5). Expanding the right side, we get y−2.25+2S=−3x−4.5−6Cx−9Cy - 2.25 + 2S = -3x - 4.5 - 6Cx - 9C. Finally, we add 2.25−2S2.25 - 2S to both sides to isolate yy and get the equation in slope-intercept form: y=−3x−4.5−6Cx−9C+2.25−2Sy = -3x - 4.5 - 6Cx - 9C + 2.25 - 2S.
  11. Combine Like Terms: With the slope of the tangent line −3−6C-3 - 6C and the point of tangency (−1.5,2.25−2S)(-1.5, 2.25 - 2S), we can use the point-slope form of the equation of a line to write the equation of the tangent line: y−(2.25−2S)=(−3−6C)(x−(−1.5))y - (2.25 - 2S) = (-3 - 6C)(x - (-1.5)). Simplifying the equation, we get y−2.25+2S=(−3−6C)(x+1.5)y - 2.25 + 2S = (-3 - 6C)(x + 1.5). Expanding the right side, we get y−2.25+2S=−3x−4.5−6Cx−9Cy - 2.25 + 2S = -3x - 4.5 - 6Cx - 9C. Finally, we add 2.25−2S2.25 - 2S to both sides to isolate yy and get the equation in slope-intercept form: y=−3x−4.5−6Cx−9C+2.25−2Sy = -3x - 4.5 - 6Cx - 9C + 2.25 - 2S. We combine like terms and use the actual values of CC and SS (rounded to three decimal places) to get the final equation of the tangent line. This will be in the form (−1.5,2.25−2S)(-1.5, 2.25 - 2S)00, where (−1.5,2.25−2S)(-1.5, 2.25 - 2S)11 is the slope and (−1.5,2.25−2S)(-1.5, 2.25 - 2S)22 is the y-intercept.

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