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The function f is defined by f(x)=x^(2)-5-5cos(3x). 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)=x2−5−5cos⁡(3x) f(x)=x^{2}-5-5 \cos (3 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:

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Q. The function f f is defined by f(x)=x2−5−5cos⁡(3x) f(x)=x^{2}-5-5 \cos (3 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 of f: To find the equation of the tangent line to the graph of f at x=−1x = -1, we need to calculate the derivative of f, which will give us the slope of the tangent line at that point.\newlineThe derivative of f with respect to xx, f′(x)f'(x), is given by:\newlinef′(x)=ddx[x2−5−5cos⁡(3x)]f'(x) = \frac{d}{dx} [x^2 - 5 - 5\cos(3x)]\newlineUsing the power rule and the chain rule, we find:\newlinef′(x)=2x−0−5⋅(−sin⁡(3x))⋅3f'(x) = 2x - 0 - 5 \cdot (-\sin(3x)) \cdot 3\newlinef′(x)=2x+15sin⁡(3x)f'(x) = 2x + 15\sin(3x)\newlineNow we need to evaluate this derivative at x=−1x = -1.\newlinef′(−1)=2(−1)+15sin⁡(3(−1))f'(-1) = 2(-1) + 15\sin(3(-1))\newlinef′(−1)=−2+15sin⁡(−3)f'(-1) = -2 + 15\sin(-3)\newlineUsing a calculator, we find sin⁡(−3)\sin(-3) and round to three decimal places.
  2. Evaluate Derivative at x=−1x = -1: Using a calculator, sin⁡(−3)≈−0.141\sin(-3) \approx -0.141. Now we plug this value into the derivative to find the slope of the tangent line at x=−1x = -1.
    f′(−1)=−2+15(−0.141)f'(-1) = -2 + 15(-0.141)
    f′(−1)=−2−15×−0.141f'(-1) = -2 - 15 \times -0.141
    f′(−1)=−2+2.115f'(-1) = -2 + 2.115
    f′(−1)≈0.115f'(-1) \approx 0.115
    This is the slope of the tangent line at x=−1x = -1.
  3. Find y-coordinate at x=−1x = -1: Next, we need to find the y-coordinate of the point on the graph of ff at x=−1x = -1 to determine the point of tangency. We do this by evaluating f(−1)f(-1).
    f(−1)=(−1)2−5−5cos⁡(3(−1))f(-1) = (-1)^2 - 5 - 5\cos(3(-1))
    f(−1)=1−5−5cos⁡(−3)f(-1) = 1 - 5 - 5\cos(-3)
    Using a calculator, we find cos⁡(−3)\cos(-3) and round to three decimal places.
  4. Determine Point of Tangency: Using a calculator, cos⁡(−3)≈0.990\cos(-3) \approx 0.990. Now we plug this value into the function to find the y-coordinate.f(−1)=1−5−5(0.990)f(-1) = 1 - 5 - 5(0.990)f(−1)=1−5−4.950f(-1) = 1 - 5 - 4.950f(−1)=−4−4.950f(-1) = -4 - 4.950f(−1)≈−8.950f(-1) \approx -8.950The point of tangency is (−1,−8.950)(-1, -8.950).
  5. Use Point-Slope Form: Now we have the slope of the tangent line, m=0.115m = 0.115, and a point on the line, (−1,−8.950)(-1, -8.950). We can use the point-slope form of the equation of a line to find the equation of the tangent line:\newliney−y1=m(x−x1)y - y_1 = m(x - x_1)\newliney−(−8.950)=0.115(x−(−1))y - (-8.950) = 0.115(x - (-1))\newliney+8.950=0.115(x+1)y + 8.950 = 0.115(x + 1)\newlineNow we simplify the equation.
  6. Simplify the Equation: Simplifying the equation, we get:\newliney+8.950=0.115x+0.115y + 8.950 = 0.115x + 0.115\newliney=0.115x+0.115−8.950y = 0.115x + 0.115 - 8.950\newliney=0.115x−8.835y = 0.115x - 8.835\newlineThis is the equation of the tangent line to the graph of ff at x=−1x = -1, rounded to three decimal places.

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