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${:[f^(')(x)=4e^(x)" and "],[f(2)=16+4e^(2).],[f(0)=◻]:}$

$\begin{array}{l}f^{\prime}(x)=4 e^{x} \text { and } f(2)=16+4 e^{2} . \\ f(0)=\square\end{array}$

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Compare linear, exponential, and quadratic growth

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Q.

\begin{array}{l}f^{\prime}(x)=4 e^{x} \text { and } f(2)=16+4 e^{2} . \\ f(0)=\square\end{array}

Understand Given Information: Understand the given information. $\newline$ We are given the derivative of a function $f(x)$ , which is $f'(x) = 4e^x$ . We are also given the value of the function at $x = 2$ , which is $f(2) = 16 + 4e^2$ . We need to find the value of the function at $x = 0$ , which is $f(0)$ .
Integrate to Find $f(x)$ : Integrate the derivative to find the general form of $f(x)$ . Since $f'(x) = 4e^x$ , we integrate to find $f(x)$ : $f(x) = \int 4e^x dx = 4e^x + C$ , where $C$ is the constant of integration.
Find Constant C: Use the given value $f(2) = 16 + 4e^2$ to find the constant C. $\newline$ We plug $x = 2$ into the general form of $f(x)$ : $\newline$ $16 + 4e^2 = 4e^2 + C$ $\newline$ Now, solve for C: $\newline$ $C = 16 + 4e^2 - 4e^2$ $\newline$ $C = 16$
Write Specific Form: Write the specific form of $f(x)$ with the found constant $C$ . Now that we have found $C$ , we can write the specific form of $f(x)$ : $f(x) = 4e^x + 16$
Calculate $f(0)$ : Calculate $f(0)$ using the specific form of $f(x)$ . $\newline$ We plug $x = 0$ into the specific form of $f(x)$ : $\newline$ $f(0) = 4e^0 + 16$ $\newline$ Since $e^0 = 1$ , we have: $\newline$ $f(0) = 4(1) + 16$ $\newline$ $f(0) = 4 + 16$ $\newline$ $f(0) = 20$

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