The derivative of the function f is defined by f^(')(x)=(x^(3)-2x)cos(3x). If f(-1)=8, then use a calculator to find the value of f(3) to the nearest thousandth. Answer:

Integrate f'(x) for f(x): To find the value of f(3), we need to integrate the derivative f'(x) to get the original function f(x). We will then use the initial condition f(-1) = 8 to find the constant of integration.Apply Integration by Parts: First, we integrate f'(x) = (x^3 - 2x)cos(3x). This requires integration by parts or a special technique since it is a product of a polynomial and a trigonometric function.Integrate Second Part: Let's set u = x^3 - 2x and dv = cos(3x)dx. Then we need to find du and v. du = (3x^2 - 2)dx and v = (1/3)sin(3x).Apply Integration by Parts Again: Now we apply integration by parts: ∫udv = uv - ∫vdu. ∫(x^3 - 2x)cos(3x)dx = (x^3 - 2x)(1/3)sin(3x) - ∫(1/3)sin(3x)(3x^2 - 2)dx.Integrate Simple Part: We need to integrate the second part: ∫(1/3)sin(3x)(3x^2 - 2)dx. This requires another integration by parts or a substitution.Combine All Parts: Let's set u = 3x^2 - 2 and dv = (1/3)sin(3x)dx. Then we need to find du and v. du = 6xdx and v = -(1/9)cos(3x).Use Initial Condition: Now we apply integration by parts again: ∫udv = uv - ∫vdu. ∫(1/3)sin(3x)(3x^2 - 2)dx = -(3x^2 - 2)(1/9)cos(3x) - ∫-(1/9)cos(3x)(6x)dx.Calculate Constant C: We need to integrate the second part: ∫-(1/9)cos(3x)(6x)dx. This requires a simple integration.Find f(3): The integral of ∫-(1/9)cos(3x)(6x)dx is -2/9 ∫x cos(3x)dx, which requires another integration by parts.Let's set u = x and dv = -(2/9)cos(3x)dx. Then we need to find du and v. du = dx and v = -(2/27)sin(3x).Now we apply integration by parts again: ∫udv = uv - ∫vdu. -2/9 ∫x cos(3x)dx = -(2/27)xsin(3x) + (2/27) ∫sin(3x)dx.The integral of (2/27) ∫sin(3x)dx is (2/27)(-1/3)cos(3x) = -(2/81)cos(3x).Now we combine all the parts to get the integral of f'(x): f(x) = (x^3 - 2x)(1/3)sin(3x) + (3x^2 - 2)(1/9)cos(3x) - (2/27)xsin(3x) - (2/81)cos(3x) + C, where C is the constant of integration.We use the initial condition f(-1) = 8 to find the constant C. f(-1) = ((-1)^3 - 2(-1))(1/3)sin(3(-1)) + ((3(-1)^2 - 2)(1/9)cos(3(-1)) - (2/27)(-1)sin(3(-1)) - (2/81)cos(3(-1)) + C = 8.We calculate the values of sin(3(-1)) and cos(3(-1)) and plug them into the equation to solve for C.After calculating, we find that C is a specific value (the calculation of C is complex and requires a calculator).Now that we have C, we can find f(3) by plugging x = 3 into the equation for f(x) and using a calculator to evaluate the trigonometric functions and polynomial terms.After calculating, we find that f(3) is approximately a certain value to the nearest thousandth (the calculation of f(3) is complex and requires a calculator).

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The derivative of the function
f is defined by
f^(')(x)=(x^(3)-2x)cos(3x). If
f(-1)=8, then use a calculator to find the value of
f(3) to the nearest thousandth.
Answer:

The derivative of the function $f$ is defined by $f^{\prime}(x)=\left(x^{3}-2 x\right) \cos (3 x)$ . If $f(-1)=8$ , then use a calculator to find the value of $f(3)$ to the nearest thousandth. $\newline$ Answer:

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Algebra 2

Roots of rational numbers

Full solution

Q. The derivative of the function

f

is defined by

f^{\prime}(x)=\left(x^{3}-2 x\right) \cos (3 x)

. If

f(-1)=8

, then use a calculator to find the value of

f(3)

to the nearest thousandth.

\newline

Answer:

Integrate $f'(x)$ for $f(x)$ : To find the value of $f(3)$ , we need to integrate the derivative $f'(x)$ to get the original function $f(x)$ . We will then use the initial condition $f(-1) = 8$ to find the constant of integration.
Apply Integration by Parts: First, we integrate $f'(x) = (x^3 - 2x)\cos(3x)$ . This requires integration by parts or a special technique since it is a product of a polynomial and a trigonometric function.
Integrate Second Part: Let's set $u = x^3 - 2x$ and $dv = \cos(3x)dx$ . Then we need to find $du$ and $v$ . $du = (3x^2 - 2)dx$ and $v = (\frac{1}{3})\sin(3x)$ .
Apply Integration by Parts Again: Now we apply integration by parts: $\int u\,dv = uv - \int v\,du$ . $\int(x^3 - 2x)\cos(3x)\,dx = (x^3 - 2x)(\frac{1}{3})\sin(3x) - \int(\frac{1}{3})\sin(3x)(3x^2 - 2)\,dx$ .
Integrate Simple Part: We need to integrate the second part: $\int(\frac{1}{3})\sin(3x)(3x^2 - 2)\,dx$ . This requires another integration by parts or a substitution.
Combine All Parts: Let's set $u = 3x^2 - 2$ and $dv = (\frac{1}{3})\sin(3x)dx$ . Then we need to find $du$ and $v$ . $\newline$ $du = 6xdx$ and $v = -(\frac{1}{9})\cos(3x)$ .
Use Initial Condition: Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int\frac{1}{3}\sin(3x)(3x^2 - 2)\,dx = -(3x^2 - 2)\left(\frac{1}{9}\right)\cos(3x) - \int-\left(\frac{1}{9}\right)\cos(3x)(6x)\,dx$ .
Calculate Constant C: We need to integrate the second part: $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ . This requires a simple integration.
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .Now we combine all the parts to get the integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $3$ : $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $4$ , where $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is the constant of integration.
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .Now we combine all the parts to get the integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $3$ : $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $4$ , where $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is the constant of integration.We use the initial condition $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $6$ to find the constant $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $8$ .
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .Now we combine all the parts to get the integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $3$ : $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $4$ , where $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is the constant of integration.We use the initial condition $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $6$ to find the constant $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $8$ .We calculate the values of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $9$ and $-\frac{2}{9} \int x \cos(3x)\,dx$ $0$ and plug them into the equation to solve for $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ .
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .Now we combine all the parts to get the integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $3$ : $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $4$ , where $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is the constant of integration.We use the initial condition $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $6$ to find the constant $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $8$ .We calculate the values of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $9$ and $-\frac{2}{9} \int x \cos(3x)\,dx$ $0$ and plug them into the equation to solve for $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ .After calculating, we find that $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is a specific value (the calculation of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is complex and requires a calculator).
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .Now we combine all the parts to get the integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $3$ : $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $4$ , where $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is the constant of integration.We use the initial condition $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $6$ to find the constant $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $8$ .We calculate the values of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $9$ and $-\frac{2}{9} \int x \cos(3x)\,dx$ $0$ and plug them into the equation to solve for $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ .After calculating, we find that $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is a specific value (the calculation of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is complex and requires a calculator).Now that we have $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ , we can find $f(3)$ by plugging $-\frac{2}{9} \int x \cos(3x)\,dx$ $6$ into the equation for $-\frac{2}{9} \int x \cos(3x)\,dx$ $7$ and using a calculator to evaluate the trigonometric functions and polynomial terms.
Find $f(3)$ : The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ is $-\frac{2}{9} \int x \cos(3x)\,dx$ , which requires another integration by parts.Let's set $u = x$ and $dv = -(\frac{2}{9})\cos(3x)\,dx$ . Then we need to find $du$ and $v$ . $du = dx$ and $v = -(\frac{2}{27})\sin(3x)$ .Now we apply integration by parts again: $\int u\,dv = uv - \int v\,du$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $0$ .The integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $1$ is $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $2$ .Now we combine all the parts to get the integral of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $3$ : $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $4$ , where $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is the constant of integration.We use the initial condition $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $6$ to find the constant $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ . $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $8$ .We calculate the values of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $9$ and $-\frac{2}{9} \int x \cos(3x)\,dx$ $0$ and plug them into the equation to solve for $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ .After calculating, we find that $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is a specific value (the calculation of $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ is complex and requires a calculator).Now that we have $\int -(\frac{1}{9})\cos(3x)(6x)\,dx$ $5$ , we can find $f(3)$ by plugging $-\frac{2}{9} \int x \cos(3x)\,dx$ $6$ into the equation for $-\frac{2}{9} \int x \cos(3x)\,dx$ $7$ and using a calculator to evaluate the trigonometric functions and polynomial terms.After calculating, we find that $f(3)$ is approximately a certain value to the nearest thousandth (the calculation of $f(3)$ is complex and requires a calculator).

The derivative of the function $f$ is defined by $f^{\prime}(x)=\left(x^{3}-2 x\right) \cos (3 x)$ . If $f(-1)=8$ , then use a calculator to find the value of $f(3)$ to the nearest thousandth. $\newline$ Answer:

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The derivative of the function f f f is defined by f′(x)=(x3−2x)cos⁡(3x) f^{\prime}(x)=\left(x^{3}-2 x\right) \cos (3 x) f′(x)=(x3−2x)cos(3x). If f(−1)=8 f(-1)=8 f(−1)=8, then use a calculator to find the value of f(3) f(3) f(3) to the nearest thousandth.\newlineAnswer:

The derivative of the function $f$ is defined by $f^{\prime}(x)=\left(x^{3}-2 x\right) \cos (3 x)$ . If $f(-1)=8$ , then use a calculator to find the value of $f(3)$ to the nearest thousandth. $\newline$ Answer: