Determine the maximum number of turning points of f.
-$$f ( x ) = \left( x + \frac { 1 } { 9 } \right) ^ { 4 } ( x - 4 ) ^ { 3 }$$
A) above the $x$-axis: $( 4 , \infty )$
below the $x$-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , \left( - \frac { 1 } { 9 } , 4 \right)$
B) above the $x$-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , \left( - \frac { 1 } { 9 } , 4 \right)$
below the $x$-axis: $( 4 , \infty )$
C) above the $x$-axis: $\left( - \frac { 1 } { 9 } , 4 \right)$
below the $x$-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , ( 4 , \infty )$
D) above the x-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , ( 4 , \infty )$
below the $x$-axis: $\left( - \frac { 1 } { 9 } , 4 \right)$

Question

Quizplus · Accepted Answer

The maximum number of turning points of a polynomial function is equal to the degree of the polynomial minus 1. The degree of the given polynomial is 7, so the maximum number of turning points is 6. The polynomial is positive for $x > 4$ and negative for $x < - \frac { 1 } { 9 }$, so it has 3 turning points above the $x$-axis and 3 turning points below the $x$-axis.

Determine the Maximum Number of Turning Points of F f(x)=(x+19)4(x−4)3f ( x ) = \left( x + \frac { 1 } { 9 } \right) ^ { 4 } ( x - 4 ) ^ { 3 }f(x)=(x+91​)4(x−4)3

Determine the Maximum Number of Turning Points of F $f ( x ) = \left( x + \frac { 1 } { 9 } \right) ^ { 4 } ( x - 4 ) ^ { 3 }$