Question

Help? Calculus

Answer 1

btw my name is michael the a and the c are just my last initials and 959 is the first number that came to mind but back to the problem

Answer 2

Well first, what are inflection points?

Answer 3

Oh, I'm sorry, Michael!

I already started it, so here's what I've done so far.

Original: f(x) = -x^4 + 24x^2 - 27
Derivative: f'(x) = -4x^3 + 48x
Second Derivative: f''(x) -12x^2 + 48

So I factored the first derivative to get: -4x(x^2 - 12) which means \[x \neq 0, \sqrt{12}\]

I plugged both of those into the second Derivative and got -96 as the max and 48 as the min, but I don't know what to do for the inflection.

Answer 4

what grade r u in?

Answer 5

Inflection points are where concavities "switch" (at least that's how I understand it personally) a point on a graph where a U will become an upside U or vice versa.

I'm in 12th grade, Michael.

Answer 6

For inflection, all we do is set the second derivative equal to 0

Why? Well what does the second derivative test give us? Concavity right? so if it is not concave up, nor down, we have an inflection point

Answer 7

Ohh. So then x = 2 (for the second derivative).

Answer 8

...Or... :P lol is there only 1 solution to that?

Answer 9

+ or - 2 because I had to take the square root!

Answer 10

Right...so we will have 2 inflection points

To find the 'y' values just plug both 2 and -2 into your original function and solve for 'y'

Answer 11

*Or just realize there is only 1 choice with those 2 as the 'x' component...but still good to do out lol

Answer 12

I got f(-2) = 53 and f(2) = 53 again

Answer 13

So, A?

Answer 14

And you would be correct....looks like an answer choice you have there :)

Answer 15

Thank you!! Can you help me out on another one if you aren't busy?

Answer 16

I got a few minutes to at least look at it lol

Answer 17

Okay ^_^