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The function 
f is defined by 
f(x)=x^(3)+1+2sin(2x). Use a calculator to write the equation of the line tangent to the graph of 
f when 
x=0.5. You should round all decimals to 3 places.
Answer:

The function f f is defined by f(x)=x3+1+2sin(2x) f(x)=x^{3}+1+2 \sin (2 x) . Use a calculator to write the equation of the line tangent to the graph of f f when x=0.5 x=0.5 . You should round all decimals to 33 places.\newlineAnswer:

Full solution

Q. The function f f is defined by f(x)=x3+1+2sin(2x) f(x)=x^{3}+1+2 \sin (2 x) . Use a calculator to write the equation of the line tangent to the graph of f f when x=0.5 x=0.5 . You should round all decimals to 33 places.\newlineAnswer:
  1. Calculate Derivative of f: To find the equation of the tangent line to the graph of ff at x=0.5x = 0.5, we need to calculate the derivative of ff, which will give us the slope of the tangent line at that point. The derivative of ff with respect to xx is f(x)=3x2+4cos(2x)f'(x) = 3x^2 + 4\cos(2x).
  2. Evaluate Derivative at x=0.5x = 0.5: Now we need to evaluate the derivative at x=0.5x = 0.5 to find the slope of the tangent line at that point. So we calculate f(0.5)=3(0.5)2+4cos(20.5)f'(0.5) = 3(0.5)^2 + 4\cos(2\cdot 0.5).
  3. Find y-coordinate at x=0.5x = 0.5: Using a calculator, we find that f(0.5)=3(0.25)+4cos(1)f'(0.5) = 3(0.25) + 4\cos(1) which is approximately 0.75+4(0.540)0.75 + 4(0.540) after rounding to three decimal places. This gives us f(0.5)0.75+2.160=2.910f'(0.5) \approx 0.75 + 2.160 = 2.910.
  4. Determine Slope and Point: Next, we need to find the yy-coordinate of the point on the graph of ff at x=0.5x = 0.5. We do this by evaluating f(0.5)=(0.5)3+1+2sin(20.5)f(0.5) = (0.5)^3 + 1 + 2\sin(2\cdot 0.5).
  5. Write Equation in Point-Slope Form: Using a calculator, we find that f(0.5)=0.125+1+2sin(1)f(0.5) = 0.125 + 1 + 2\sin(1) which is approximately 0.125+1+2(0.841)0.125 + 1 + 2(0.841) after rounding to three decimal places. This gives us f(0.5)0.125+1+1.682=2.807f(0.5) \approx 0.125 + 1 + 1.682 = 2.807.
  6. Convert to Slope-Intercept Form: Now we have the slope of the tangent line, m=2.910m = 2.910, and a point on the tangent line, (0.5,2.807)(0.5, 2.807). We can use the point-slope form of a line to write the equation of the tangent line: yy1=m(xx1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is the point on the line.
  7. Final Equation: Substituting the values we have, the equation of the tangent line is y2.807=2.910(x0.5)y - 2.807 = 2.910(x - 0.5).
  8. Final Equation: Substituting the values we have, the equation of the tangent line is y2.807=2.910(x0.5)y - 2.807 = 2.910(x - 0.5).To write the equation in slope-intercept form, we distribute the slope on the right side and add 2.8072.807 to both sides: y=2.910x1.455+2.807y = 2.910x - 1.455 + 2.807.
  9. Final Equation: Substituting the values we have, the equation of the tangent line is y2.807=2.910(x0.5)y - 2.807 = 2.910(x - 0.5).To write the equation in slope-intercept form, we distribute the slope on the right side and add 2.8072.807 to both sides: y=2.910x1.455+2.807y = 2.910x - 1.455 + 2.807.Simplifying the equation, we get y=2.910x+1.352y = 2.910x + 1.352 after rounding to three decimal places.

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