Question

Let f(x)=1/(8x^2+6). I need to find the intervals on which the function is concave up and where it is down. I keep getting imaginary numbers so it doesn't make sense. I need assistance

Answer 1

whats your 2nd derivative show you?

Answer 2

f'(x) = -16x
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(8x^2 +6)^2

Answer 3

at x = 0 there is a critical point, and probably and inflection.... better consult the 2nd derivative for concavity

Answer 4

f''(x) = (8x^2 +6)^2 (-16) - (-16x)(2)(8x^2 +6)(16x)
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[(8x^2 +6)^2]^2

Answer 5

(0+6)^2 (-16) - 0(......) = 36(-16) = (-)

it looks like x=0 is a Maximum value.....

Answer 6

our inflections will be when that top of the 2nd derivative =0

Answer 7

this making any sense to you?

Answer 8

Yeah I got the x=o part and that it was a maximum. But why is the top of the second derivative equaling zero the inflection point. I thought when the second derivative equals zero is your point of inflection. That would mean that the bottom equals zero.

Answer 9

the 2nd derivative = 0 indicates that the concavity of the graph changes at that point, it is no longer cave up or cave down at "0"; it is switching concavities.

So the top of your 2nd derivative = 0 will produce the f''(x) = 0 effect... the bottom is just there for decoration ;)

Answer 10

lets clean this top up so we can use it..

(8x^2 +6)^2 (-16) - (-16x)(2)(8x^2 +6)(16x)

(-16)(64x^4 +96x^2 +36) + 512x^2(8x^2 +6)

Answer 11

+4096x^4 -1024x^4 + 3072x^2 -1536x^2 -576

Answer 12

3072x^4 + 1536x^2 -576 you see any common factors :)

Answer 13

this is a disguised quadratic; a^2 = x^4; a = x^2

3072a^2 + 1536a -576

Answer 14

okay i think i can take it from here. Thank you oh so much :D

Answer 15

no problem; just be aware that any value you get for the new variable "a" needs to be equated to a^2 = x^4 and a = x^2... dont just think that "a" is the answer half way thru ;) youll do fine....

Answer 16

One last question. If i use x=\[-b\pm\[\sqrt{?}\]b^2-4ac \div\]2a
are the answers for x squared
or can i even use that equation

Answer 17

I meant the quadratic formula

Answer 18

you can use that equation to get your answers for "a"...and then plug that value for "a" into the:

x^4 = a^2

and

x^2 = a

equations.

but I am not sure how well the quad formula works when you have common factors still in it, I think it might give you a false reading:

try it on a simpler equation to find out like:

2x^2 +12x +18.... that is just 2(x^2 +6x +9) so if the answers are the same, your good to go, but if they are different, yo need to factor out the common number ;)

Answer 19

We are good to go!

Answer 20

yay!! ..simplicity at work lol