{:[f^(')(x)=-5e^(x)" and "],[f(3)=22-5e^(3).],[f(0)=◻]:}

Integrate f'(x): To find f(0), we need to integrate the derivative f'(x) to get the original function f(x). The given derivative is f'(x) = -5e^x. Integration of f'(x) will give us f(x) plus a constant C, which we can determine using the given f(3) = 22 - 5e^3. Let's integrate f'(x) = -5e^x. ∫(-5e^x) dx = -5∫e^x dx = -5e^x + CFind Constant C: Now we need to find the value of the constant C using the given f(3) = 22 - 5e^3. We substitute x with 3 in the integrated function f(x) = -5e^x + C. -5e^3 + C = 22 - 5e^3 Now, we solve for C. C = 22 - 5e^3 + 5e^3 C = 22Calculate f(0): With the value of C found, we can now write the complete function f(x): f(x) = -5e^x + 22 To find f(0), we substitute x with 0 in the function f(x). f(0) = -5e^0 + 22We know that e^0 is 1, so we can simplify f(0) as follows: f(0) = -5(1) + 22 f(0) = -5 + 22 f(0) = 17

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Calculus

Find higher derivatives of rational and radical functions

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\begin{array}{l}f^{\prime}(x)=-5 e^{x} \text { and } f(3)=22-5 e^{3} . \\ f(0)=\square\end{array}

Integrate $f'(x)$ : To find $f(0)$ , we need to integrate the derivative $f'(x)$ to get the original function $f(x)$ . The given derivative is $f'(x) = -5e^x$ . $\newline$ Integration of $f'(x)$ will give us $f(x)$ plus a constant $C$ , which we can determine using the given $f(3) = 22 - 5e^3$ . $\newline$ Let's integrate $f'(x) = -5e^x$ . $\newline$ $f(0)$ $0$
Find Constant C: Now we need to find the value of the constant $C$ using the given $f(3) = 22 - 5e^3$ . We substitute $x$ with $3$ in the integrated function $f(x) = -5e^x + C$ . $-5e^3 + C = 22 - 5e^3$ Now, we solve for $C$ . $C = 22 - 5e^3 + 5e^3$ $C = 22$
Calculate $f(0)$ : With the value of $C$ found, we can now write the complete function $f(x)$ : $f(x) = -5e^x + 22$ To find $f(0)$ , we substitute $x$ with $0$ in the function $f(x)$ . $f(0) = -5e^0 + 22$
Calculate $f(0)$ : With the value of $C$ found, we can now write the complete function $f(x)$ : $f(x) = -5e^x + 22$ To find $f(0)$ , we substitute $x$ with $0$ in the function $f(x)$ . $f(0) = -5e^0 + 22$ We know that $e^0$ is $1$ , so we can simplify $f(0)$ as follows: $f(0) = -5(1) + 22$ $f(0) = -5 + 22$ $f(0) = 17$