The function f is defined by f(x)=x^(3)-2cos(2x^(2)+2). 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. Answer:

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The function  f is defined by  f(x)=x^(3)-2cos(2x^(2)+2). 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. Answer:

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Calculate Derivative: To find the equation of the tangent line at x = 0.5, we first need to calculate the derivative of f(x), which will give us the slope of the tangent line at any point x. The derivative of f(x) = x^3 - 2cos(2x^2 + 2) is f'(x) = 3x^2 + 4xsin(2x^2 + 2).Find Slope at x=0.5: Next, we evaluate the derivative at x = 0.5 to find the slope of the tangent line at that point. f'(0.5) = 3(0.5)^2 + 4(0.5)sin(2(0.5)^2 + 2).Find y-coordinate at x=0.5: Performing the calculations, we get: f'(0.5) = 3(0.25) + 2sin(2(0.25) + 2) = 0.75 + 2sin(2.5). Using a calculator, we find sin(2.5) ≈ -0.598. So, f'(0.5) ≈ 0.75 + 2(-0.598) ≈ 0.75 - 1.196 ≈ -0.446. We round this to three decimal places, getting f'(0.5) ≈ -0.446.Write Point-Slope Equation: Now we need to find the y-coordinate of the function at x = 0.5 to get the point through which the tangent line passes. f(0.5) = (0.5)^3 - 2cos(2(0.5)^2 + 2) = 0.125 - 2cos(3). Using a calculator, we find cos(3) ≈ -0.990. So, f(0.5) ≈ 0.125 - 2(-0.990) ≈ 0.125 + 1.980 ≈ 2.105. We round this to three decimal places, getting f(0.5) ≈ 2.105.Convert to Slope-Intercept Form: With the slope of the tangent line and the point (0.5, 2.105), we can use the point-slope form of the equation of a line to write the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line. Substituting our values, we get y - 2.105 = -0.446(x - 0.5).To write the equation in slope-intercept form, we simplify the equation: y = -0.446x + 0.223 + 2.105. Combining like terms, we get y = -0.446x + 2.328. We round the constants to three decimal places, so the final equation is y ≈ -0.446x + 2.328.

The function $f$ is defined by $f(x)=x^{3}-2 \cos \left(2 x^{2}+2\right)$ . 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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The function f f f is defined by f(x)=x3−2cos⁡(2x2+2) f(x)=x^{3}-2 \cos \left(2 x^{2}+2\right) f(x)=x3−2cos(2x2+2). Use a calculator to write the equation of the line tangent to the graph of f f f when x=0.5 x=0.5 x=0.5. You should round all decimals to 333 places.\newlineAnswer:

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