The function $f$ is defined by $f(x)=x^{3}-2 \sin \left(x^{2}\right)$ . 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^{3}-2 \sin \left(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: To find the equation of the tangent line at $x = 1$ , we need to calculate the derivative of $f(x)$ to find the slope of the tangent line at that point. The derivative $f'(x)$ will be calculated using the power rule for $x^3$ and the chain rule for $-2\sin(x^2)$ .
Find Slope: First, we find the derivative of $x^3$ , which is $3x^2$ . Then, we find the derivative of $-2\sin(x^2)$ . The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ , and by the chain rule, the derivative of $u = x^2$ is $2x$ . So, the derivative of $-2\sin(x^2)$ is $-2\cos(x^2) \cdot 2x = -4x\cos(x^2)$ .
Evaluate at $x=1$ : Now we combine the derivatives to get the derivative of $f(x)$ : $f'(x) = 3x^2 - 4x\cos(x^2)$ .
Calculate y-coordinate: Next, we evaluate $f'(x)$ at $x = 1$ to find the slope of the tangent line at that point: $f'(1) = 3(1)^2 - 4(1)\cos(1^2) = 3 - 4\cos(1)$ . We use a calculator to find the value of $\cos(1)$ and then calculate the slope.
Use Point-Slope Form: Using a calculator, we find that $\cos(1)$ is approximately $0.540$ . So, $f'(1) \approx 3 - 4(0.540) = 3 - 2.160 = 0.840$ . This is the slope of the tangent line at $x = 1$ .
Find Tangent Line Equation: We also need the $y$ -coordinate of the point on the graph of $f(x)$ at $x = 1$ to find the equation of the tangent line. We calculate $f(1) = 1^3 - 2\sin(1^2) = 1 - 2\sin(1)$ . Again, we use a calculator to find the value of $\sin(1)$ .
Find Tangent Line Equation: We also need the y-coordinate of the point on the graph of $f(x)$ at $x = 1$ to find the equation of the tangent line. We calculate $f(1) = 1^3 - 2\sin(1^2) = 1 - 2\sin(1)$ . Again, we use a calculator to find the value of $\sin(1)$ .Using a calculator, we find that $\sin(1)$ is approximately $0.841$ . So, $f(1) \approx 1 - 2(0.841) = 1 - 1.682 = -0.682$ . This is the y-coordinate of the point on the graph of $f(x)$ at $x = 1$ .
Find Tangent Line Equation: We also need the $y$ -coordinate of the point on the graph of $f(x)$ at $x = 1$ to find the equation of the tangent line. We calculate $f(1) = 1^3 - 2\sin(1^2) = 1 - 2\sin(1)$ . Again, we use a calculator to find the value of $\sin(1)$ .Using a calculator, we find that $\sin(1)$ is approximately $0.841$ . So, $f(1) \approx 1 - 2(0.841) = 1 - 1.682 = -0.682$ . This is the $y$ -coordinate of the point on the graph of $f(x)$ at $x = 1$ .Now we have the slope of the tangent line, $f(x)$ $1$ , and a point on the line, $f(x)$ $2$ . We can use the point-slope form of the equation of a line, $f(x)$ $3$ , to find the equation of the tangent line.
Find Tangent Line Equation: We also need the y-coordinate of the point on the graph of $f(x)$ at $x = 1$ to find the equation of the tangent line. We calculate $f(1) = 1^3 - 2\sin(1^2) = 1 - 2\sin(1)$ . Again, we use a calculator to find the value of $\sin(1)$ .Using a calculator, we find that $\sin(1)$ is approximately $0.841$ . So, $f(1) \approx 1 - 2(0.841) = 1 - 1.682 = -0.682$ . This is the y-coordinate of the point on the graph of $f(x)$ at $x = 1$ .Now we have the slope of the tangent line, $m = 0.840$ , and a point on the line, $x = 1$ $0$ . We can use the point-slope form of the equation of a line, $x = 1$ $1$ , to find the equation of the tangent line.Plugging in the values, we get the equation of the tangent line: $x = 1$ $2$ . Simplifying, we get $x = 1$ $3$ .
Find Tangent Line Equation: We also need the $y$ -coordinate of the point on the graph of $f(x)$ at $x = 1$ to find the equation of the tangent line. We calculate $f(1) = 1^3 - 2\sin(1^2) = 1 - 2\sin(1)$ . Again, we use a calculator to find the value of $\sin(1)$ .Using a calculator, we find that $\sin(1)$ is approximately $0.841$ . So, $f(1) \approx 1 - 2(0.841) = 1 - 1.682 = -0.682$ . This is the $y$ -coordinate of the point on the graph of $f(x)$ at $x = 1$ .Now we have the slope of the tangent line, $f(x)$ $1$ , and a point on the line, $f(x)$ $2$ . We can use the point-slope form of the equation of a line, $f(x)$ $3$ , to find the equation of the tangent line.Plugging in the values, we get the equation of the tangent line: $f(x)$ $4$ . Simplifying, we get $f(x)$ $5$ .Finally, we write the equation in slope-intercept form, $f(x)$ $6$ , by isolating $y$ : $f(x)$ $8$ . Combining the constants, we get $f(x)$ $9$ .