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The function 
f is defined by 
f(x)=x^(3)+3-5sin(x^(2)). 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+35sin(x2) f(x)=x^{3}+3-5 \sin \left(x^{2}\right) . 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)=x3+35sin(x2) f(x)=x^{3}+3-5 \sin \left(x^{2}\right) . 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)=x3+35sin(x2)f(x) = x^3 + 3 - 5\sin(x^2) using the power rule and the chain rule is:\newlinef(x)=3x210xcos(x2)f'(x) = 3x^2 - 10x \cdot \cos(x^2).\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)=3(1)210(1)cos(12)=310cos(1)f'(1) = 3(1)^2 - 10(1) \cdot \cos(1^2) = 3 - 10\cos(1). Using a calculator, we find that cos(1)\cos(1) is approximately 0.5400.540 (rounded to three decimal places). So, f(1)3100.540=35.4=2.4f'(1) \approx 3 - 10 \cdot 0.540 = 3 - 5.4 = -2.4. The slope of the tangent line at x=1x = 1 is approximately 2.4-2.4.
  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. This is done by evaluating the original function at x=1x = 1.f(1)=13+35sin(12)=1+35sin(1)f(1) = 1^3 + 3 - 5\sin(1^2) = 1 + 3 - 5\sin(1).Using a calculator, we find that sin(1)\sin(1) is approximately 0.8410.841 (rounded to three decimal places).So, f(1)1+35×0.841=44.205=0.205f(1) \approx 1 + 3 - 5 \times 0.841 = 4 - 4.205 = -0.205.The y-coordinate of the point on the graph at x=1x = 1 is approximately 0.205-0.205.
  4. Use Point-Slope Form: Now we have the slope of the tangent line 2.4-2.4 and a point on the line (1,0.205)(1, -0.205). We can use the point-slope form of the equation of a line to find the equation of the tangent line:\newlineyy1=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.\newlinePlugging in our values, we get:\newliney(0.205)=2.4(x1)y - (-0.205) = -2.4(x - 1).
  5. Simplify and Finalize Equation: Simplifying the equation, we get:\newliney+0.205=2.4x+2.4y + 0.205 = -2.4x + 2.4.\newlineSubtracting 0.2050.205 from both sides to get yy by itself, we have:\newliney=2.4x+2.40.205y = -2.4x + 2.4 - 0.205.\newlineCombining like terms, we get:\newliney=2.4x+2.195y = -2.4x + 2.195.\newlineThis is the equation of the tangent line, rounded to three decimal places.

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