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${:[f^(')(x)=12e^(x)" and "f(4)=-16+12e^(4).],[f(0)=]:}$

$\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}$

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Find higher derivatives of rational and radical functions

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Q.

\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}

Integrate $f'(x)$ : To find the value of $f(x)$ at $x = 0$ , we need to integrate the derivative $f'(x)$ to get $f(x)$ . The derivative $f'(x) = 12e^x$ is given. $\newline$ Integration of $f'(x)$ will give us $f(x) = 12e^x + C$ , where $C$ is the constant of integration.
Find Constant of Integration: We are given the value of the function at $x = 4$ , which is $f(4) = -16 + 12e^4$ . We can use this information to find the constant of integration $C$ . Let's substitute $x = 4$ into the integrated function $f(x) = 12e^x + C$ . $12e^4 + C = -16 + 12e^4$
Calculate Complete Function: Solving for $C$ , we subtract $12e^4$ from both sides of the equation: $\newline$ $C = -16 + 12e^4 - 12e^4$ $\newline$ $C = -16$
Substitute $x = 0$ : Now that we have the constant of integration, we can write the complete function $f(x)$ : $f(x) = 12e^x - 16$
Simplify Expression: To find $f(0)$ , we substitute $x = 0$ into the function $f(x)$ : $f(0) = 12e^{0} - 16$
Simplify Expression: To find $f(0)$ , we substitute $x = 0$ into the function $f(x)$ :
$f(0) = 12e^0 - 16$ Since $e^0$ is equal to $1$ , we can simplify the expression:
$f(0) = 12(1) - 16$
$f(0) = 12 - 16$
$f(0) = -4$

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