The function $f$ is defined by $f(x)=x^{2}+4 x+3 \cos \left(2 x^{2}\right)$ . 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}+4 x+3 \cos \left(2 x^{2}\right)

. 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 of f: To find the equation of the tangent line to the graph of $f$ at $x = 1$ , we need to calculate the derivative of $f$ at $x = 1$ , which will give us the slope of the tangent line. The derivative of $f$ , denoted as $f'(x)$ , is found using the power rule for $x^2$ and $4x$ , and the chain rule for $3\cos(2x^2)$ .
Combine and Evaluate Derivative: First, we differentiate $x^2$ to get $2x$ , and $4x$ to get $4$ . For $3\cos(2x^2)$ , we use the chain rule: the derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ , and the derivative of $2x^2$ with respect to $x$ is $4x$ . So, the derivative of $3\cos(2x^2)$ is $2x$ $2$ .
Find y-coordinate at $x=1$ : Now we combine the derivatives: $f'(x) = 2x + 4 - 12x\sin(2x^2)$ . We then evaluate this derivative at $x = 1$ to find the slope of the tangent line.
Use Point-Slope Form: Plugging $x = 1$ into the derivative, we get $f'(1) = 2(1) + 4 - 12(1)\sin(2(1)^2) = 2 + 4 - 12\sin(2)$ . We use a calculator to find $\sin(2)$ and round to three decimal places.
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .Next, we need to find the y-coordinate of the point on the graph of $f$ at $x = 1$ . We do this by evaluating $f(1) = 1^2 + 4(1) + 3\cos(2(1)^2) = 1 + 4 + 3\cos(2)$ .
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .Next, we need to find the y-coordinate of the point on the graph of $f$ at $x = 1$ . We do this by evaluating $f(1) = 1^2 + 4(1) + 3\cos(2(1)^2) = 1 + 4 + 3\cos(2)$ .Using the calculator again, $\cos(2)$ is approximately $0.540$ . So, $f(1) = 1 + 4 + 3 \times 0.540 \approx 1 + 4 + 1.620 = 5 + 1.620 = 6.620$ . This is the y-coordinate of the point of tangency.
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .Next, we need to find the y-coordinate of the point on the graph of $f$ at $x = 1$ . We do this by evaluating $f(1) = 1^2 + 4(1) + 3\cos(2(1)^2) = 1 + 4 + 3\cos(2)$ .Using the calculator again, $\cos(2)$ is approximately $0.540$ . So, $f(1) = 1 + 4 + 3 \times 0.540 \approx 1 + 4 + 1.620 = 5 + 1.620 = 6.620$ . This is the y-coordinate of the point of tangency.Now we have the slope of the tangent line, $0.909$ $0$ , and a point on the line, $0.909$ $1$ . We can use the point-slope form of a line, $0.909$ $2$ , where $0.909$ $3$ is the slope and $0.909$ $4$ is the point on the line.
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .Next, we need to find the y-coordinate of the point on the graph of $f$ at $x = 1$ . We do this by evaluating $f(1) = 1^2 + 4(1) + 3\cos(2(1)^2) = 1 + 4 + 3\cos(2)$ .Using the calculator again, $\cos(2)$ is approximately $0.540$ . So, $f(1) = 1 + 4 + 3 \times 0.540 \approx 1 + 4 + 1.620 = 5 + 1.620 = 6.620$ . This is the y-coordinate of the point of tangency.Now we have the slope of the tangent line, $0.909$ $0$ , and a point on the line, $0.909$ $1$ . We can use the point-slope form of a line, $0.909$ $2$ , where $0.909$ $3$ is the slope and $0.909$ $4$ is the point on the line.Plugging in our values, we get the equation of the tangent line: $0.909$ $5$ . To write this in slope-intercept form, we distribute the slope and simplify.
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .Next, we need to find the y-coordinate of the point on the graph of $f$ at $x = 1$ . We do this by evaluating $f(1) = 1^2 + 4(1) + 3\cos(2(1)^2) = 1 + 4 + 3\cos(2)$ .Using the calculator again, $\cos(2)$ is approximately $0.540$ . So, $f(1) = 1 + 4 + 3 \times 0.540 \approx 1 + 4 + 1.620 = 5 + 1.620 = 6.620$ . This is the y-coordinate of the point of tangency.Now we have the slope of the tangent line, $0.909$ $0$ , and a point on the line, $0.909$ $1$ . We can use the point-slope form of a line, $0.909$ $2$ , where $0.909$ $3$ is the slope and $0.909$ $4$ is the point on the line.Plugging in our values, we get the equation of the tangent line: $0.909$ $5$ . To write this in slope-intercept form, we distribute the slope and simplify.After distributing, we get $0.909$ $6$ . Adding $0.909$ $7$ to both sides to solve for $0.909$ $8$ , we get $0.909$ $9$ .
Write Equation in Slope-Intercept Form: Using a calculator, $\sin(2)$ is approximately $0.909$ . So, $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ . This is the slope of the tangent line at $x = 1$ .Next, we need to find the y-coordinate of the point on the graph of $f$ at $x = 1$ . We do this by evaluating $f(1) = 1^2 + 4(1) + 3\cos(2(1)^2) = 1 + 4 + 3\cos(2)$ .Using the calculator again, $\cos(2)$ is approximately $0.540$ . So, $f(1) = 1 + 4 + 3 \times 0.540 \approx 1 + 4 + 1.620 = 5 + 1.620 = 6.620$ . This is the y-coordinate of the point of tangency.Now we have the slope of the tangent line, $0.909$ $0$ , and a point on the line, $0.909$ $1$ . We can use the point-slope form of a line, $0.909$ $2$ , where $0.909$ $3$ is the slope and $0.909$ $4$ is the point on the line.Plugging in our values, we get the equation of the tangent line: $0.909$ $5$ . To write this in slope-intercept form, we distribute the slope and simplify.After distributing, we get $0.909$ $6$ . Adding $0.909$ $7$ to both sides to solve for $0.909$ $8$ , we get $0.909$ $9$ .Combining like terms, we get $f'(1) = 2 + 4 - 12 \times 0.909 \approx 2 + 4 - 10.908 = 6 - 10.908 \approx -4.908$ $0$ . This is the equation of the tangent line to the graph of $f$ at $x = 1$ , rounded to three decimal places.