Find $$ f $$ such that $$ f^{\prime}(x)=4 x^{2}+3 x-3 $$ and $$ f(0)=7 $$ $$ f(x)=\square $$

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$$ f(x) = \frac{4}{3}x^3 + \frac{3}{2}x^2 - 3x + 7 $$ To find the function $$ f(x) $$ such that $$ f'(x) = 4x^2 + 3x - 3 $$ and $$ f(0) = 7 $$, we need to integrate the derivative $$ f'(x) $$. 1. **Integrate $$ f'(x) $$**: $$ f(x) = \int (4x^2 + 3x - 3) \, dx $$ We can integrate each term separately: - The integral of $$ 4x^2 $$ is $$ \frac{4}{3}x^3 $$. - The integral of $$ 3x $$ is $$ \frac{3}{2}x^2 $$. - The integral of $$ -3 $$ is $$ -3x $$. Therefore, we have: $$ f(x) = \frac{4}{3}x^3 + \frac{3}{2}x^2 - 3x + C $$ where $$ C $$ is the constant of integration. 2. **Use the initial condition $$ f(0) = 7 $$**: Substitute $$ x = 0 $$ into the equation: $$ f(0) = \frac{4}{3}(0)^3 + \frac{3}{2}(0)^2 - 3(0) + C = C $$ Since $$ f(0) = 7 $$, we have: $$ C = 7 $$ 3. **Write the final expression for $$ f(x) $$**: Substitute $$ C $$ back into the equation for $$ f(x) $$: $$ f(x) = \frac{4}{3}x^3 + \frac{3}{2}x^2 - 3x + 7 $$ Thus, the function $$ f(x) $$ is: $$ \boxed{\frac{4}{3}x^3 + \frac{3}{2}x^2 - 3x + 7} $$

Find \( f \) such that \( f^{\prime}(x)=4 x^{2}+3 x-3 \) and \( f(0)=7 \) \( f(x)=\square \)

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Find \( f \) such that \( f^{\prime}(x)=4 x^{2}+3 x-3 \) and \( f(0)=7 \) \( f(x)=\square \)

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