For what value(s) of x does f(x)=−1x2+1 have an inflection point? x=±5√5 x = ± 5 5 x=±13 x = ± 1 3 x=±3√3 x = ± 3 3 x=0 x = 0 x=0, −1 x = 0 , − 1
To find the possible inflection points, we need to find the values of x where the second derivative of the function is equal to zero or does not exist.
First, let's find the second derivative of f(x):
f(x) = -x^2 + 1
f'(x) = -2x
f''(x) = -2
Since the second derivative is a constant (-2), it is never equal to zero. Therefore, there are NO inflection points for the given function f(x) = -x^2 + 1.
Therefore, none of the options listed are the correct answer.