Question

Find the critical and inflection points of the graph

Answer 1

\[\frac{x^2}{(x-6)^2}\]

Answer 2

I have \[f \prime (x) = \frac{-12x}{(x-6)^3}\] so the C.N would be 0 right?

Answer 3

6 wouldnt be a C.N because its not in the domain correct??

Answer 4

Yeah something seems to be happening at x = 0,

what do you mean it's not in the domain?

Answer 5

I understood that a critical point has to be in the domain of the original function

Answer 6

0 is in the domain of f(x)

Answer 7

right.. a critical number is where the derivative is also undefined.. The derivative is undefined at 6 but 6 isnt in the domain of the original function

Answer 8

hmm... right.

Answer 9

wait that's wrong. there is a critical point at x = 0

Answer 10

it seems to be a local minimum, since f''(x) | x = 0 is a positive number.

Answer 11

yeah zero is a cn

Answer 12

so there is a local minimum at (0,0)

to find the points of inflection, solve f''(x) = 0

Answer 13

\[f \prime \prime (x) = \frac{24(x+3)}{(x-6)^4}\]

Answer 14

I found a point of inflection at x = -3

so there is a point of inflection at (-3, 1/9)

Answer 15

Cn at -3?

Answer 16

there's no critical number at x = -3. there's a point of inflection

Answer 17

Could you help me with the concave up and concave down please??

Answer 18

Yeah i ment inflection sorry lol

Answer 19

I guess the function is concave down before x = -3, and concave up after x = -3