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Find the range of the function $f(x),$ where
\[f(x) = x^k\]
for $x \ge 2$, and $k = -3$.

 Aug 17, 2023
 #1
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Given the function \(f(x) = x^k\) with \(x \ge 2\) and \(k = -3\), let's find its range.

For \(x \ge 2\) and \(k = -3\), the function becomes \(f(x) = x^{-3}\).

The range of a function is the set of all possible output values it can produce. Since the domain is \(x \ge 2\), let's examine how the function behaves as \(x\) increases:

- As \(x\) increases beyond 2, \(x^{-3}\) becomes smaller and smaller. As \(x\) approaches infinity, \(x^{-3}\) approaches 0.

Since \(x^{-3}\) approaches 0 as \(x\) increases, the range of \(f(x) = x^{-3}\) is \((0, \infty)\), which means all positive real numbers excluding zero. The function cannot produce negative values or zero because it is defined only for \(x \ge 2\).

Therefore, the range of the function \(f(x) = x^k\) with \(x \ge 2\) and \(k = -3\) is \((0, \infty)\).

 Aug 17, 2023

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