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

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Find equations of tangent lines using limits

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Q. The function

f

is defined by

f(x)=x^{2}+x-2 \sin (2 x)

. Use a calculator to write the equation of the line tangent to the graph of

f

when

x=3

. You should round all decimals to

3

places.

\newline

Answer:

Calculate Derivative and Slope: To find the equation of the tangent line at $x = 3$ , we first need to calculate the derivative of the function $f(x)$ to find the slope of the tangent line at that point. $\newline$ The derivative of $f(x) = x^2 + x - 2\sin(2x)$ is $f'(x) = 2x + 1 - 4\cos(2x)$ . $\newline$ Now we will calculate $f'(3)$ to get the slope of the tangent line at $x = 3$ .
Calculate $f'(3)$ : Using a calculator, we find the value of $f'(3) = 2(3) + 1 - 4\cos(2\times 3)$ . This gives us $f'(3) = 6 + 1 - 4\cos(6)$ . Now we will calculate the numerical value of $\cos(6)$ and then find the value of $f'(3)$ .
Find y-coordinate at x= $3$ : Assuming the calculator is set to radian mode, we find that $\cos(6) \approx 0.960$ (rounded to three decimal places). $\newline$ Now we calculate $f'(3) = 6 + 1 - 4(0.960)$ . $\newline$ This gives us $f'(3) = 7 - 3.840 = 3.160$ (rounded to three decimal places). $\newline$ The slope of the tangent line at $x = 3$ is $3.160$ .
Use Point-Slope Form: Next, we need to find the $y$ -coordinate of the point on the graph of $f(x)$ at $x = 3$ to use it in the point-slope form of the equation of the tangent line. $\newline$ We calculate $f(3) = 3^2 + 3 - 2\sin(2\cdot 3)$ .
Write Equation in Slope-Intercept Form: Using the calculator, we find that $\sin(6) \approx -0.279$ (rounded to three decimal places). $\newline$ Now we calculate $f(3) = 9 + 3 - 2(-0.279)$ . $\newline$ This gives us $f(3) = 12 + 0.558 = 12.558$ (rounded to three decimal places). $\newline$ The y-coordinate of the point on the graph of $f(x)$ at $x = 3$ is $12.558$ .
Write Equation in Slope-Intercept Form: Using the calculator, we find that $\sin(6) \approx -0.279$ (rounded to three decimal places). Now we calculate $f(3) = 9 + 3 - 2(-0.279)$ . This gives us $f(3) = 12 + 0.558 = 12.558$ (rounded to three decimal places). The $y$ -coordinate of the point on the graph of $f(x)$ at $x = 3$ is $12.558$ .We now have the slope of the tangent line ( $m = 3.160$ ) and a point on the tangent line $(3, 12.558)$ . Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$ , we can write the equation of the tangent line. Substituting the values, we get $f(3) = 9 + 3 - 2(-0.279)$ $0$ .
Write Equation in Slope-Intercept Form: Using the calculator, we find that $\sin(6) \approx -0.279$ (rounded to three decimal places). Now we calculate $f(3) = 9 + 3 - 2(-0.279)$ . This gives us $f(3) = 12 + 0.558 = 12.558$ (rounded to three decimal places). The $y$ -coordinate of the point on the graph of $f(x)$ at $x = 3$ is $12.558$ .We now have the slope of the tangent line ( $m = 3.160$ ) and a point on the tangent line $(3, 12.558)$ . Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$ , we can write the equation of the tangent line. Substituting the values, we get $f(3) = 9 + 3 - 2(-0.279)$ $0$ .To write the equation in slope-intercept form, we simplify the equation: $f(3) = 9 + 3 - 2(-0.279)$ $1$ . Expanding this, we get $f(3) = 9 + 3 - 2(-0.279)$ $2$ . Finally, we combine like terms to get $f(3) = 9 + 3 - 2(-0.279)$ $3$ . This is the equation of the tangent line to the graph of $f(x)$ at $x = 3$ , rounded to three decimal places.