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${:[f^(')(x)=-(4)/(x^(2))" and "f(2)=4],[f(1)=]:}$

$\begin{array}{l}f^{\prime}(x)=-\frac{4}{x^{2}} \text { and } f(2)=4 \\ f(1)=\end{array}$

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Find derivatives of logarithmic functions

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Q.

\begin{array}{l}f^{\prime}(x)=-\frac{4}{x^{2}} \text { and } f(2)=4 \\ f(1)=\end{array}

Understand the problem: Understand the problem. $\newline$ We are given the derivative of a function $f(x)$ , which is $f'(x) = -\frac{4}{x^2}$ , and a point on the function, $f(2) = 4$ . We need to find the value of the function at $x = 1$ , which is $f(1)$ .
Integrate the derivative: Integrate the derivative to find the original function. $\newline$ To find $f(x)$ , we need to integrate $f'(x)$ . The integral of $f'(x) = -\frac{4}{x^2}$ with respect to x is: $\newline$ $\int f'(x) \, dx = \int -\frac{4}{x^2} \, dx = 4\int -\frac{1}{x^2} \, dx = 4\int x^{-2} \, dx$ $\newline$ The antiderivative of $x^{-2}$ is $-x^{-1}$ , so we have: $\newline$ $f(x) = 4(-x^{-1}) + C$ $\newline$ $f(x) = -\frac{4}{x} + C$ , where $C$ is the constant of integration.
Use the given point: Use the given point to find the constant of integration. $\newline$ We know that $f(2) = 4$ , so we can plug this into the equation to solve for $C$ : $\newline$ $4 = -\frac{4}{2} + C$ $\newline$ $4 = -2 + C$ $\newline$ $C = 4 + 2$ $\newline$ $C = 6$ $\newline$ Now we have the original function: $\newline$ $f(x) = -\frac{4}{x} + 6$
Find $f(1)$ : Find $f(1)$ using the original function. $\newline$ Now that we have $f(x)$ , we can find $f(1)$ by plugging in $x = 1$ : $\newline$ $f(1) = -\frac{4}{1} + 6$ $\newline$ $f(1) = -4 + 6$ $\newline$ $f(1) = 2$

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