The function $f$ is defined by $f(x)=x^{2}+\sin (2 x)$ . 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. $\newline$ 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}+\sin (2 x)

. 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.

\newline

Answer:

Calculate Derivative and Slope: To find the equation of the tangent line at $x = -1$ , we first need to calculate the derivative of the function $f(x) = x^2 + \sin(2x)$ , which will give us the slope of the tangent line at any point $x$ . Using the power rule and the chain rule for derivatives, we get: $f'(x) = 2x + \cos(2x) \cdot 2$ Now we need to evaluate this derivative at $x = -1$ .
Evaluate Derivative at $x = -1$ : Plugging $x = -1$ into the derivative, we get: $\newline$ $f'(-1) = 2(-1) + \cos(2(-1)) \cdot 2$ $\newline$ $f'(-1) = -2 + \cos(-2) \cdot 2$ $\newline$ Using a calculator, we find that $\cos(-2)$ is approximately $0.416$ (rounded to three decimal places). $\newline$ So, $f'(-1) = -2 + 0.416 \cdot 2$ $\newline$ $f'(-1) = -2 + 0.832$ $\newline$ $f'(-1) = -1.168$ $\newline$ This is the slope of the tangent line at $x = -1$ .
Find Point of Tangency: Next, we need to find the $y$ -coordinate of the point on the graph of $f(x)$ at $x = -1$ to determine the point through which the tangent line passes. $\newline$ We calculate $f(-1) = (-1)^2 + \sin(2(-1))$ $\newline$ $f(-1) = 1 + \sin(-2)$ $\newline$ Using a calculator, we find that $\sin(-2)$ is approximately $-0.909$ (rounded to three decimal places). $\newline$ So, $f(-1) = 1 - 0.909$ $\newline$ $f(-1) = 0.091$ $\newline$ The point of tangency is $(-1, 0.091)$ .
Write Equation of Tangent Line: Now we have a point $(-1, 0.091)$ and a slope $-1.168$ for the tangent line. We can use the point-slope form of a line to write the equation of the tangent line: $\newline$ $y - y_1 = m(x - x_1)$ $\newline$ $y - 0.091 = -1.168(x - (-1))$ $\newline$ $y - 0.091 = -1.168(x + 1)$
Write Equation of Tangent Line: Now we have a point $(-1, 0.091)$ and a slope $-1.168$ for the tangent line. We can use the point-slope form of a line to write the equation of the tangent line: $\newline$ $y - y_1 = m(x - x_1)$ $\newline$ $y - 0.091 = -1.168(x - (-1))$ $\newline$ $y - 0.091 = -1.168(x + 1)$ Finally, we simplify the equation to get it into slope-intercept form $(y = mx + b)$ : $\newline$ $y = -1.168x - 1.168 + 0.091$ $\newline$ $y = -1.168x - 1.077$ $\newline$ This is the equation of the tangent line, rounded to three decimal places.