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Describe the end behavior of \( f(x)=-x^{7} \) As \( x \rightarrow-\infty, f(x) \rightarrow \) ? As \( x \rightarrow \infty, f(x) \rightarrow \) ? Describe the end behavior of \( g(x)=-x^{4} \) As \( x \rightarrow-\infty, g(x) \rightarrow \) ? ?

Ask by Collins Morgan. in the United States
Mar 17,2025

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For \( f(x)=-x^7 \): - As \( x \) approaches \(-\infty\), \( f(x) \) approaches \(\infty\). - As \( x \) approaches \(\infty\), \( f(x) \) approaches \(-\infty\). For \( g(x)=-x^4 \): - As \( x \) approaches \(-\infty\), \( g(x) \) approaches \(-\infty\).

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When we look at the function \( f(x) = -x^7 \), we see that as \( x \rightarrow -\infty \), the value of \( f(x) \) will approach \( +\infty \) because raising a negative number to an odd power results in a negative value, and the negative sign in front turns that positive. Conversely, as \( x \rightarrow \infty \), \( f(x) \) will approach \( -\infty \), since a positive number raised to an odd power is negative due to the negative sign in front. Turning our attention to \( g(x) = -x^4 \), we observe that as \( x \rightarrow -\infty \), \( g(x) \) will approach \( -\infty \) because raising a negative number to an even power yields a positive number, and again, the negative sign flips that positive result to negative. Meanwhile, as \( x \rightarrow \infty \), \( g(x) \) will also approach \( -\infty \) for the same reason—an even powered term remains positive but is negated. It’s a dramatic downward trend either way!

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