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${:[f^(')(x)=-3e^(x)" and "],[f(1)=12-3e.],[f(0)=◻]:}$

$\begin{array}{l}f^{\prime}(x)=-3 e^{x} \text { and } f(1)=12-3 e . \\ f(0)=\square\end{array}$

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Find derivatives of sine and cosine functions

Full solution

Q.

\begin{array}{l}f^{\prime}(x)=-3 e^{x} \text { and } f(1)=12-3 e . \\ f(0)=\square\end{array}

Integrate $f'(x)$ : To find $f(0)$ , we need to integrate the derivative $f'(x) = -3e^x$ to get $f(x)$ . Let's integrate $f'(x)$ to find the general form of $f(x)$ . $\newline$ $\int f'(x) \, dx = \int (-3e^x) \, dx$ $\newline$ $f(x) = -3\int e^x \, dx$ $\newline$ $f(x) = -3e^x + C$ , where $C$ is the constant of integration.
Find Constant C: Now we need to find the value of the constant $C$ using the given condition $f(1) = 12 - 3e$ . Let's substitute $x = 1$ into $f(x)$ to find $C$ . $f(1) = -3e^1 + C = 12 - 3e$ $C = 12 - 3e + 3e$ $C = 12$
Write Complete Function: With the value of $C$ found, we can now write the complete function $f(x)$ : $f(x) = -3e^x + 12$
Find $f(0)$ : Finally, we can find $f(0)$ by substituting $x = 0$ into $f(x)$ :
$f(0) = -3e^0 + 12$
$f(0) = -3(1) + 12$
$f(0) = -3 + 12$
$f(0) = 9$

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