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The function f is defined by f(x)=x^(2)+cos(3x^(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)=x2+cos⁡(3x2) f(x)=x^{2}+\cos \left(3 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:

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Q. The function f f is defined by f(x)=x2+cos⁡(3x2) f(x)=x^{2}+\cos \left(3 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. Find Derivative of Function: To find the equation of the tangent line at x=−1x = -1, we first need to find the derivative of the function f(x)=x2+cos⁡(3x2)f(x) = x^2 + \cos(3x^2), which will give us the slope of the tangent line at any point xx.
  2. Calculate Derivative at x=−1x = -1: The derivative of f(x)f(x) with respect to xx is f′(x)=2x−sin⁡(3x2)⋅6xf'(x) = 2x - \sin(3x^2) \cdot 6x. This uses the chain rule for the cosine term, where the derivative of cos⁡(u)\cos(u) with respect to xx is −sin⁡(u)⋅dudx-\sin(u) \cdot \frac{du}{dx}, and u=3x2u = 3x^2.
  3. Find y-coordinate at x=−1x = -1: Now we need to evaluate the derivative at x=−1x = -1 to find the slope of the tangent line at that point. So we calculate f′(−1)=2(−1)−sin⁡(3(−1)2)⋅6(−1)f'(-1) = 2(-1) - \sin(3(-1)^2) \cdot 6(-1).
  4. Use Point-Slope Form: Plugging in the values, we get f′(−1)=−2−sin⁡(3)×−6f'(-1) = -2 - \sin(3) \times -6. Using a calculator, we find sin⁡(3)\sin(3) and round it to three decimal places.
  5. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).
  6. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the y-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).
  7. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the y-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).Plugging in the values, we get f(−1)=1+cos⁡(3)f(-1) = 1 + \cos(3). Using a calculator, we find cos⁡(3)\cos(3) and round it to three decimal places.
  8. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the yy-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).Plugging in the values, we get f(−1)=1+cos⁡(3)f(-1) = 1 + \cos(3). Using a calculator, we find cos⁡(3)\cos(3) and round it to three decimal places.Using a calculator, cos⁡(3)≈−0.990\cos(3) \approx -0.990, so f(−1)=1−0.990=0.010f(-1) = 1 - 0.990 = 0.010 (rounded to three decimal places).
  9. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the y-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).Plugging in the values, we get f(−1)=1+cos⁡(3)f(-1) = 1 + \cos(3). Using a calculator, we find cos⁡(3)\cos(3) and round it to three decimal places.Using a calculator, cos⁡(3)≈−0.990\cos(3) \approx -0.990, so f(−1)=1−0.990=0.010f(-1) = 1 - 0.990 = 0.010 (rounded to three decimal places).Now we have the slope of the tangent line, m=−1.154m = -1.154, and a point on the line, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15400. We can use the point-slope form of a line to write the equation: f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15411.
  10. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the y-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).Plugging in the values, we get f(−1)=1+cos⁡(3)f(-1) = 1 + \cos(3). Using a calculator, we find cos⁡(3)\cos(3) and round it to three decimal places.Using a calculator, cos⁡(3)≈−0.990\cos(3) \approx -0.990, so f(−1)=1−0.990=0.010f(-1) = 1 - 0.990 = 0.010 (rounded to three decimal places).Now we have the slope of the tangent line, m=−1.154m = -1.154, and a point on the line, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15400. We can use the point-slope form of a line to write the equation: f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15411.Substituting the values into the point-slope form, we get f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15422.
  11. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the y-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).Plugging in the values, we get f(−1)=1+cos⁡(3)f(-1) = 1 + \cos(3). Using a calculator, we find cos⁡(3)\cos(3) and round it to three decimal places.Using a calculator, cos⁡(3)≈−0.990\cos(3) \approx -0.990, so f(−1)=1−0.990=0.010f(-1) = 1 - 0.990 = 0.010 (rounded to three decimal places).Now we have the slope of the tangent line, m=−1.154m = -1.154, and a point on the line, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15400. We can use the point-slope form of a line to write the equation: f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15411.Substituting the values into the point-slope form, we get f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15422.Finally, we simplify the equation to get it into slope-intercept form, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15433. So, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15444.
  12. Convert to Slope-Intercept Form: Using a calculator, sin⁡(3)≈0.141\sin(3) \approx 0.141, so f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.154 (rounded to three decimal places).Next, we need to find the yy-coordinate of the function f(x)f(x) at x=−1x = -1 to find the point of tangency. We calculate f(−1)=(−1)2+cos⁡(3(−1)2)f(-1) = (-1)^2 + \cos(3(-1)^2).Plugging in the values, we get f(−1)=1+cos⁡(3)f(-1) = 1 + \cos(3). Using a calculator, we find cos⁡(3)\cos(3) and round it to three decimal places.Using a calculator, cos⁡(3)≈−0.990\cos(3) \approx -0.990, so f(−1)=1−0.990=0.010f(-1) = 1 - 0.990 = 0.010 (rounded to three decimal places).Now we have the slope of the tangent line, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15400, and a point on the line, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15411. We can use the point-slope form of a line to write the equation: f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15422.Substituting the values into the point-slope form, we get f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15433.Finally, we simplify the equation to get it into slope-intercept form, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15444. So, f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15455.Simplifying the yy-intercept, we get f′(−1)=−2−(0.141×−6)=−2+0.846=−1.154f'(-1) = -2 - (0.141 \times -6) = -2 + 0.846 = -1.15477. This is the equation of the tangent line at x=−1x = -1, rounded to three decimal places.

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