Question

please explain

Answer 1

what is the point?

Answer 2

@Loser66 knows this really well :P

Answer 3

Your question: "Does this tangent line lie above or below the graph at this point"
My question: what is the point?

Answer 4

is it (3,4)??

Answer 5

I admire mondona, he never gives up. hehehe... He makes this question more than 3times. :)

Answer 6

lol?? Perseverance!

Answer 7

find the second derivative

Answer 8

\[g'(x) = \dfrac{x^2-16}{x-2}\]

\[g''(x) = ?\]

Answer 9

knock knock @mondona

Answer 10

how ? you need to use quotient rule to find the derivative

Answer 11

doesnt look correct, try again and show all steps so that i can pinpoint the mistake

Answer 12

right, next step

Answer 13

Do you know why you are calculating the second derivative?

Answer 14

\[g'(x) = \dfrac{x^2-16}{x-2}\]

\[g''(x) = \dfrac{\frac{d}{dx}(x^2-16) (x-2) - \frac{d}{dx}(x-2)(x^2-16)}{(x-2)^2}\]

\[g''(x) = \dfrac{(2x) (x-2) - (1)(x^2-16)}{(x-2)^2}\]

yes ?

Answer 15

distribute and simplify the numerator

Answer 16

\[g''(x) = \dfrac{2x^2-4x - x^2+16}{(x-2)^2}\]

Answer 17

\[g''(x) = \dfrac{x^2-4x+16}{(x-2)^2}\]

Answer 18

still yes ?

Answer 19

good, find g''(3)

Answer 20

plugin x=3 above

Answer 21

which is positive

Answer 22

g''(3) > 0

so can we say the graph of g(x) is concave up (valley) at x=3 ?