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${:[f^(')(x)=-7e^(x)" and "],[f(5)=24-7e^(5).],[f(0)=◻]:}$

$\begin{array}{l}f^{\prime}(x)=-7 e^{x} \text { and } f(5)=24-7 e^{5} . \\ f(0)=\square\end{array}$

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Power rule with rational exponents

Full solution

Q.

\begin{array}{l}f^{\prime}(x)=-7 e^{x} \text { and } f(5)=24-7 e^{5} . \\ f(0)=\square\end{array}

Derivative of $f(x)$ : We have $f'(x) = -7e^x$ , which is the derivative of $f(x)$ . To find $f(0)$ , we need to integrate $f'(x)$ .
Integration of $f'(x)$ : The integral of $f'(x) = -7e^x$ is $f(x) = -7e^x + C$ , where $C$ is the constant of integration.
Finding Constant of Integration: We know $f(5) = 24 - 7e^5$ . Let's plug $x = 5$ into the integrated function to find $C$ .
Substitute $x = 5$ : $-7e^5 + C = 24 - 7e^5$ .
Solving for C: Solving for C, we get $C = 24$ .
Final Function $f(x)$ : Now we have the function $f(x) = -7e^x + 24$ .
Finding $f(0)$ : To find $f(0)$ , we plug in $x = 0$ into $f(x)$ .
Substitute $x = 0$ : $f(0) = -7e^0 + 24$ .
Calculate $f(0)$ : Since $e^0 = 1$ , $f(0) = -7(1) + 24$ .
Result: $f(0) = -7 + 24$ .
Result: $f(0) = -7 + 24$ . $f(0) = 17$ .

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