Determine the x values that cause the function to be (a) zero, (b) undefined, (c) positive, and (d) negative.
-$$f ( x ) = \frac { ( 2 x + 1 ) \sqrt { x - 2 } } { ( x - 8 ) ^ { 2 } }$$
A) (a) $\{ 2 , - 1 \}$, (b) $\{ 8 \} \cup ( \infty , 2 )$, (c) $( 2,8 ) \cup ( 8 , \infty )$, (d) $\varnothing$
B) (a) $\{ 2 \}$, (b) $\{ 8 \} \cup ( \infty , 2 )$, (c) $\varnothing$, (d) $( 2,8 ) \cup ( 8 , \infty )$
C) (a) $\{ 2 \}$, (b) $\{ 8 \} \cup ( \infty , 2 )$, (c) $( 2,8 ) \cup ( 8 , \infty )$, (d) $\varnothing$
D) (a) $\{ 2 , - 1 \}$, (b) $\{ 8 \} \cup ( \infty , 2 )$, (c) $( 2,8 )$, (d) $( 8 , \infty )$

Question

Quizplus · Accepted Answer

(a) The function is zero when the numerator is zero, which happens at $x = -\frac{1}{2}$ (not $-1$) and not at $x = 2$ since $\sqrt{x - 2}$ is zero at $x = 2$, making the whole expression zero. (b) The function is undefined at $x = 8$ due to division by zero and for $x < 2$ because the square root of a negative number is undefined in real numbers. (c) The function is positive for $x > 2$ and $x \neq 8$ because both the numerator and denominator are positive for $x > 2$ (excluding $x = 8$), and the square root function is defined and positive. (d) There are no values of $x$ for which the function is negative because the square root and the squared terms ensure the function's output is always non-negative for its domain.

Determine the X Values That Cause the Function to Be f(x)=(2x+1)x−2(x−8)2f ( x ) = \frac { ( 2 x + 1 ) \sqrt { x - 2 } } { ( x - 8 ) ^ { 2 } }f(x)=(x−8)2(2x+1)x−2​​

Determine the X Values That Cause the Function to Be $f ( x ) = \frac { ( 2 x + 1 ) \sqrt { x - 2 } } { ( x - 8 ) ^ { 2 } }$