$f ( x ) = \frac { x ^ { 2 } + 1 } { x ^ { 2 } }$
A) (1,2)
B) No turning points
C) (-1,-2)
D) (0,0)

Name: $f ( x ) = \frac { x ^ { 2 } + 1 } { x ^ { 2 } }$ A) (1,2) B) No turning points C) (-1,-2) D) (0,0)
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Question

$f ( x ) = \frac { x ^ { 2 } + 1 } { x ^ { 2 } }$
A)  (1,2) 
B) No turning points
C)  (-1,-2) 
D)  (0,0)

Quizplus · Accepted Answer

The function $f(x) = \frac{x^2 + 1}{x^2}$ simplifies to $f(x) = 1 + \frac{1}{x^2}$, which has no real roots and its derivative $f'(x) = -\frac{2}{x^3}$ does not change sign, indicating no turning points.

f(x)=x2+1x2f ( x ) = \frac { x ^ { 2 } + 1 } { x ^ { 2 } }f(x)=x2x2+1​

$f ( x ) = \frac { x ^ { 2 } + 1 } { x ^ { 2 } }$