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f^(')(x)=(100)/(x^(3)) and
f(5)=-14.

f(1)=

$f^{\prime}(x)=\frac{100}{x^{3}}$ and $f(5)=-14$ . $\newline$ $f(1)=$

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Operations with rational exponents

Full solution

Q.

f^{\prime}(x)=\frac{100}{x^{3}}

and

f(5)=-14

.

\newline

f(1)=

Integrate $f'(x)$ for $f(x)$ : To find $f(1)$ , we need to integrate $f'(x)$ to get $f(x)$ . The integral of $f'(x) = \frac{100}{x^3}$ is $f(x) = -\frac{50}{x^2} + C$ , where $C$ is the constant of integration.
Find Constant C: We know $f(5) = -14$ . Let's plug $x = 5$ into $f(x)$ to find $C$ .
$-14 = -\frac{50}{(5^2)} + C$
$-14 = -\frac{50}{25} + C$
$-14 = -2 + C$
$C = -14 + 2$
$C = -12$
Calculate $f(1)$ : Now we have $f(x) = -\frac{50}{x^2} - 12$ . Let's find $f(1)$ by plugging in $x = 1$ .
$f(1) = -\frac{50}{(1^2)} - 12$
$f(1) = -50 - 12$
$f(1) = -62$

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