The function $f$ is defined by $f(x)=x^{3}-2 \sin \left(x^{2}+2 x\right)$ . Use a calculator to write the equation of the line tangent to the graph of $f$ when $x=0.5$ . 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^{3}-2 \sin \left(x^{2}+2 x\right)

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

f

when

x=0.5

. You should round all decimals to

3

places.

\newline

Answer:

Calculate Derivative and Slope: To find the equation of the tangent line at $x = 0.5$ , we need to calculate the derivative of $f(x)$ to find the slope of the tangent line at that point. $\newline$ The derivative of $f(x) = x^3 - 2\sin(x^2 + 2x)$ is $f'(x) = 3x^2 - 2(2x + 2)\cos(x^2 + 2x)$ . $\newline$ Now we need to evaluate this derivative at $x = 0.5$ .
Evaluate Derivative at $x = 0.5$ : Using a calculator, we find the value of $f'(0.5) = 3(0.5)^2 - 2(2 \times 0.5 + 2)\cos((0.5)^2 + 2 \times 0.5)$ . This simplifies to $f'(0.5) = 3(0.25) - 2(1 + 2)\cos(0.25 + 1)$ .
Find y-coordinate at x = $0$ . $5$ : Now we calculate the numerical value of $f'(0.5)$ using a calculator. $\newline$ $f'(0.5) \approx 0.75 - 2(3)\cos(1.25)$ . $\newline$ $f'(0.5) \approx 0.75 - 6\cos(1.25)$ . $\newline$ After using a calculator, we get $f'(0.5) \approx 0.75 - 6 \times \cos(1.25) \approx 0.75 - 6 \times 0.315$ . $\newline$ $f'(0.5) \approx 0.75 - 1.89$ . $\newline$ $f'(0.5) \approx -1.14$ (rounded to three decimal places).
Calculate Tangent Line Equation: Next, we need to find the y-coordinate of the function $f(x)$ at $x = 0.5$ to use it as a point on the tangent line. $\newline$ We calculate $f(0.5) = (0.5)^3 - 2\sin((0.5)^2 + 2\cdot0.5)$ . $\newline$ This simplifies to $f(0.5) = 0.125 - 2\sin(0.25 + 1)$ .
Calculate Tangent Line Equation: Next, we need to find the y-coordinate of the function $f(x)$ at $x = 0.5$ to use it as a point on the tangent line. $\newline$ We calculate $f(0.5) = (0.5)^3 - 2\sin((0.5)^2 + 2\cdot0.5)$ . $\newline$ This simplifies to $f(0.5) = 0.125 - 2\sin(0.25 + 1)$ .Using a calculator, we find the value of $f(0.5) \approx 0.125 - 2\sin(1.25)$ . $\newline$ After using a calculator, we get $f(0.5) \approx 0.125 - 2 \cdot \sin(1.25) \approx 0.125 - 2 \cdot 0.948$ . $\newline$ $f(0.5) \approx 0.125 - 1.896$ . $\newline$ $f(0.5) \approx -1.771$ (rounded to three decimal places).
Calculate Tangent Line Equation: Next, we need to find the y-coordinate of the function $f(x)$ at $x = 0.5$ to use it as a point on the tangent line. $\newline$ We calculate $f(0.5) = (0.5)^3 - 2\sin((0.5)^2 + 2\cdot0.5)$ . $\newline$ This simplifies to $f(0.5) = 0.125 - 2\sin(0.25 + 1)$ .Using a calculator, we find the value of $f(0.5) \approx 0.125 - 2\sin(1.25)$ . $\newline$ After using a calculator, we get $f(0.5) \approx 0.125 - 2 \cdot \sin(1.25) \approx 0.125 - 2 \cdot 0.948$ . $\newline$ $f(0.5) \approx 0.125 - 1.896$ . $\newline$ $f(0.5) \approx -1.771$ (rounded to three decimal places).Now we have the slope of the tangent line, which is $f'(0.5) \approx -1.14$ , and a point on the line, which is $(0.5, -1.771)$ . $\newline$ We can use the point-slope form of the equation of a line to write the equation of the tangent line: $x = 0.5$ $0$ . $\newline$ Substituting the values, we get $x = 0.5$ $1$ .
Calculate Tangent Line Equation: Next, we need to find the y-coordinate of the function $f(x)$ at $x = 0.5$ to use it as a point on the tangent line. $\newline$ We calculate $f(0.5) = (0.5)^3 - 2\sin((0.5)^2 + 2\cdot0.5)$ . $\newline$ This simplifies to $f(0.5) = 0.125 - 2\sin(0.25 + 1)$ .Using a calculator, we find the value of $f(0.5) \approx 0.125 - 2\sin(1.25)$ . $\newline$ After using a calculator, we get $f(0.5) \approx 0.125 - 2 \cdot \sin(1.25) \approx 0.125 - 2 \cdot 0.948$ . $\newline$ $f(0.5) \approx 0.125 - 1.896$ . $\newline$ $f(0.5) \approx -1.771$ (rounded to three decimal places).Now we have the slope of the tangent line, which is $f'(0.5) \approx -1.14$ , and a point on the line, which is $(0.5, -1.771)$ . $\newline$ We can use the point-slope form of the equation of a line to write the equation of the tangent line: $x = 0.5$ $0$ . $\newline$ Substituting the values, we get $x = 0.5$ $1$ .Simplifying the equation, we get $x = 0.5$ $2$ . $\newline$ Then, $x = 0.5$ $3$ . $\newline$ $x = 0.5$ $4$ (rounded to three decimal places). $\newline$ This is the equation of the tangent line at $x = 0.5$ .