$\begin{array}{l}f^{\prime}(x)=5 e^{x} \text { and } f(7)=40+5 e^{7} . \\ f(0)=\square\end{array}$

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Find higher derivatives of rational and radical functions

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\begin{array}{l}f^{\prime}(x)=5 e^{x} \text { and } f(7)=40+5 e^{7} . \\ f(0)=\square\end{array}

Given Derivative and Function Value: We are given the derivative of a function $f(x)$ , which is $f'(x) = 5e^x$ . We also know the value of the function at $x = 7$ , which is $f(7) = 40 + 5e^7$ . To find $f(0)$ , we need to integrate the derivative $f'(x)$ to get the original function $f(x)$ and then evaluate it at $x = 0$ .
Integrating the Derivative: The integral of $f'(x) = 5e^x$ with respect to $x$ is $f(x) = 5e^x + C$ , where $C$ is the constant of integration.
Finding the Constant: To find the constant $C$ , we use the known value of $f(7) = 40 + 5e^7$ . We substitute $x = 7$ into the integrated function and set it equal to $40 + 5e^7$ : $5e^7 + C = 40 + 5e^7.$
Substituting $x = 0$ : Solving for $C$ , we subtract $5e^7$ from both sides of the equation: $\newline$ $C = 40 + 5e^7 - 5e^7$ .
Calculating $f(0)$ : This simplifies to $C = 40$ , since $5e^7 - 5e^7 = 0$ .
Calculating $f(0)$ : This simplifies to $C = 40$ , since $5e^7 - 5e^7 = 0$ .Now we have the original function $f(x) = 5e^x + 40$ . To find $f(0)$ , we substitute $x = 0$ into the function: $\newline$ $f(0) = 5e^0 + 40$ .
Calculating $f(0)$ : This simplifies to $C = 40$ , since $5e^7 - 5e^7 = 0$ .Now we have the original function $f(x) = 5e^x + 40$ . To find $f(0)$ , we substitute $x = 0$ into the function: $\newline$ $f(0) = 5e^0 + 40$ .Since $e^0 = 1$ , the equation simplifies to: $\newline$ $f(0) = 5(1) + 40$ .
Calculating $f(0)$ : This simplifies to $C = 40$ , since $5e^7 - 5e^7 = 0$ .Now we have the original function $f(x) = 5e^x + 40$ . To find $f(0)$ , we substitute $x = 0$ into the function: $\newline$ $f(0) = 5e^0 + 40$ .Since $e^0 = 1$ , the equation simplifies to: $\newline$ $f(0) = 5(1) + 40$ .Calculating the value, we get: $\newline$ $f(0) = 5 + 40$ .
Calculating $f(0)$ : This simplifies to $C = 40$ , since $5e^7 - 5e^7 = 0$ .Now we have the original function $f(x) = 5e^x + 40$ . To find $f(0)$ , we substitute $x = 0$ into the function: $\newline$ $f(0) = 5e^0 + 40$ .Since $e^0 = 1$ , the equation simplifies to: $\newline$ $f(0) = 5(1) + 40$ .Calculating the value, we get: $\newline$ $f(0) = 5 + 40$ .This gives us the final answer: $\newline$ $C = 40$ $0$ .