Question

directions: for each of the following, determine if the given polynomial must have a global minimum, global maximum, neither. explain your reasoning.
(15) $f(x)=x^{4}-5x^{3}+x + 6$
mini:
maxi:
(16) $y=-2x^{3}-x^{2}+8x$
(17) $g(x)=-x^{6}+x^{3}+4x^{2}+1$

Explanation:

Step1: Recall polynomial - end - behavior rules

For a polynomial \(y = a_nx^n+\cdots+a_0\), if \(n\) is even and \(a_n>0\), the function has a global minimum. If \(n\) is even and \(a_n < 0\), the function has a global maximum. If \(n\) is odd, the function has neither a global maximum nor a global minimum.

Step2: Analyze \(f(x)=x^{4}-5x^{3}+x + 6\)

The degree \(n = 4\) (even) and the leading - coefficient \(a_n=1>0\). As \(x\to\pm\infty\), \(y\to+\infty\). So, \(f(x)\) has a global minimum.

Step3: Analyze \(y=-2x^{3}-x^{2}+8x\)

The degree \(n = 3\) (odd). As \(x\to+\infty\), \(y\to-\infty\) and as \(x\to-\infty\), \(y\to+\infty\). So, \(y\) has neither a global maximum nor a global minimum.

Step4: Analyze \(g(x)=-x^{6}+x^{3}+4x^{2}+1\)

The degree \(n = 6\) (even) and the leading - coefficient \(a_n=-1<0\). As \(x\to\pm\infty\), \(y\to-\infty\). So, \(g(x)\) has a global maximum.

Answer:

1. Global minimum. Reason: Degree is 4 (even) and leading - coefficient is positive.
2. Neither. Reason: Degree is 3 (odd).
3. Global maximum. Reason: Degree is 6 (even) and leading - coefficient is negative.