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

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

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Find higher derivatives of rational and radical functions

Full solution

Q.

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

Substitute $x$ with $4$ : To find $f(4)$ , substitute $x$ with $4$ in the function $f(x) = \frac{16}{x^2}$ . $\newline$ $f(4) = \frac{16}{4^2}$
Calculate $4^2$ : Calculate the value of $4^2$ . $\newline$ $4^2 = 16$
Divide $16$ by $16$ : Now, divide $16$ by $16$ . $\newline$ $f(4) = \frac{16}{16}$
Simplify the division: Simplify the division to get the final answer. $f(4) = 1$

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