point of inflection

[maeveoneill]

when determining the points of inflection, they were the same as my critical points. does that mean that poitns of inflection do not exist or am i wrong?

my equation is (x^2 -4)/ x^2 -1)

thank you!

 

[pinkie]

well a point of inflection occurs when ur second derivative is equal to zero. If at that certain point u dont have a max or a min, than its a point of inflection. It cant be a max/min and a point of inflection at the same time. ♥

 

[maeveoneill]

okay so it doesnt mean that everything is wrong. it just means that ther are no points of inflection? thank you

if anyone wanted to be a help.. i gathered that my vertical asymptope is at x=1 and x=-1
horizontal asymtop[e at y=1
x intercepts are +2, -2
y intercept is 4
critical points are (2,1.3)(-2,-1.3) and x=+1, -1 but that is the vertical asympotope.

if someone could check that over for me i would love you

 

[Unco]

i gathered that my vertical asymptote is at x=1 and x=-1 Good
horizontal asymtop[e at y=1 Good
x intercepts are +2, -2 Yep
y intercept is 4 Yep

critical points are (2,1.3)(-2,-1.3) and x=+1, -1 but that is the vertical asymptote.

\(\displaystyle \L \mbox{f(x) = \frac{x^2 - 4}{x^2 - 1}}\)

Applying the quotient rule and simplifying gives

\(\displaystyle \L \mbox{f'(x) = \frac{6x}{(x^2 - 1)^2}}\)

\(\displaystyle \mbox{x = \pm 1}\) make \(\displaystyle \mbox{f'(x)}\) undefined, not zero. \(\displaystyle \mbox{x = 0}\) is the only solution to \(\displaystyle \mbox{f'(x) = 0}\).

Applying the quotient rule again and simplifying gives
\(\displaystyle \L \mbox{f''(x) = \frac{-6(3x^2 + 1)}{(x^2 - 1)^3}}\)

Is this ever going to be zero (for x is real)?

Conclusion:

 

[maeveoneill]

thank you.. actaully critical points would be at x =0, +1 and -1 because critical points exist where f'x =0 and whre f'x does not exist. but iw as using the wrong equation for wehre f'x =0. so thanks.

 

[Unco]

Darn, you got me!