Please Help!

1/8x^2+6 Find concave up and concave down intervals also inflection point

Question

Please Help!

1/8x^2+6 Find concave up and concave down intervals also inflection point

anonymous · Answer

I found f'=-16x/(8x^2+6)^2
f'' not sure if this is right but i got (-1024x^4+2560x^2+2496)/(8x^2+6)^4

campbell_st · Answer

well thats ok.... now solve f'(x) = 0

this will only occur when x = 0

that's the only stationary point

test it in the 2nd derivative and you get 

and I think there is an error in the 2nd derivative

I found 

$$f'(x) = -16x(8x^2 +6)^{-2}$$

and before simplifying... I used the product rule rather then quotient 

$$f''(x) = \frac{512x^2 -128x - 96}{(8x^2 + 6)^3}$$


so testing the stationary point x = 0

f''(0) = -96/6^3

less than zero so x = 0 is a max... 

the denominator will always be positive so no vertical asymptotes 

and a horizontal asymptote at x = 0

substitute it into the original equation and you'll find that the max value is 1/6

hope it helps

campbell_st · Answer

so since you only have a max... it concave down... I suppose.. 

it looks like

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