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The function 
f is defined by 
f(x)=x^(2)+5+3sin(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)=x2+5+3sin(2x) f(x)=x^{2}+5+3 \sin (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)=x2+5+3sin(2x) f(x)=x^{2}+5+3 \sin (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: To find the equation of the tangent line at x=1x = 1, we first need to calculate the derivative of the function f(x)f(x) to find the slope of the tangent line at that point.\newlineThe derivative of f(x)=x2+5+3sin(2x)f(x) = x^2 + 5 + 3\sin(2x) is f(x)=2x+3×2cos(2x)=2x+6cos(2x)f'(x) = 2x + 3 \times 2\cos(2x) = 2x + 6\cos(2x).\newlineNow we need to evaluate this derivative at x=1x = 1.
  2. Evaluate Derivative at x=1x=1: Evaluating the derivative at x=1x = 1 gives us f(1)=2(1)+6cos(2(1))=2+6cos(2)f'(1) = 2(1) + 6\cos(2(1)) = 2 + 6\cos(2). Using a calculator, we find that cos(2)\cos(2) is approximately 0.5400.540. So, f(1)2+6×0.540=2+3.24=5.24f'(1) \approx 2 + 6 \times 0.540 = 2 + 3.24 = 5.24. We round this to three decimal places, so the slope of the tangent line at x=1x = 1 is approximately 5.2405.240.
  3. Find y-coordinate at x=11: Next, we need to find the y-coordinate of the point on the graph of f(x)f(x) where x=1x = 1. We do this by evaluating the original function at x=1x = 1: f(1)=12+5+3sin(21)=1+5+3sin(2)f(1) = 1^2 + 5 + 3\sin(2\cdot1) = 1 + 5 + 3\sin(2). Using a calculator, we find that sin(2)\sin(2) is approximately 0.9090.909. So, f(1)1+5+30.909=6+2.727=8.727f(1) \approx 1 + 5 + 3 \cdot 0.909 = 6 + 2.727 = 8.727. We round this to three decimal places, so the y-coordinate is approximately 8.7278.727.
  4. Write Point-Slope Equation: Now we have the slope of the tangent line m=5.240m = 5.240 and a point on the tangent line x=1,y8.727x = 1, y \approx 8.727. 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). Substituting the known values, we get y8.727=5.240(x1)y - 8.727 = 5.240(x - 1).
  5. Convert to Slope-Intercept Form: To write the equation in slope-intercept form y=mx+by = mx + b, we simplify the equation from the previous step:\newliney=5.240(x1)+8.727y = 5.240(x - 1) + 8.727\newliney=5.240x5.240+8.727y = 5.240x - 5.240 + 8.727\newliney=5.240x+3.487y = 5.240x + 3.487\newlineWe round the y-intercept to three decimal places, so the equation of the tangent line is y5.240x+3.487y \approx 5.240x + 3.487.

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