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If
f(x)=(-x^(2)-2)/(7x), what is the value of
f(8), to the nearest tenth (if necessary)?
Answer:

If $f(x)=\frac{-x^{2}-2}{7 x}$ , what is the value of $f(8)$ , to the nearest tenth (if necessary)? $\newline$ Answer:

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Transformations of functions

Full solution

Q. If

f(x)=\frac{-x^{2}-2}{7 x}

, what is the value of

f(8)

, to the nearest tenth (if necessary)?

\newline

Answer:

Substitute $x$ with $8$ : Substitute the value of $x$ with $8$ in the function $f(x)$ . We have $f(x) = \frac{-x^2 - 2}{7x}$ , so $f(8) = \frac{-(8)^2 - 2}{7 \times 8}$ .
Calculate numerator for $x=8$ : Calculate the numerator of the function when $x$ is $8$ . The numerator is $-\left(8\right)^2 - 2$ , which is $-64 - 2 = -66$ .
Calculate denominator for $x=8$ : Calculate the denominator of the function when $x$ is $8$ . The denominator is $7 \times 8$ , which is $56$ .
Divide numerator by denominator: Divide the numerator by the denominator to find $f(8)$ . So, $f(8) = \frac{-66}{56}$ .
Simplify fraction for $f(8)$ : Simplify the fraction to get the value of $f(8)$ to the nearest tenth. $\frac{-66}{56}$ simplifies to approximately $-1.1786$ , which rounds to $-1.2$ when rounded to the nearest tenth.

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