Question

PLEASE HELPP

Answer 1

Hahaha... I saw you posted the problem 3 times, whyzzzzz???

Answer 2

this is another part to it, but i have still not got a response for the other one.. nobodys responding

Answer 3

ok, take steps:
derivative of g is g', and it is \(g'=\dfrac{x^2-16}{x-2}\), the graph gets critical value when g' =0, and we know that g' =0 iff x^2-16 =0 or x = -4 or x =4
get it?

Answer 4

so, critical value is at x =4, and x =-4

Answer 5

not sure yet, but we have to find which one is max/ min.

Answer 6

Let me work out on paper.

Answer 7

ok, both are min values

Answer 8

do you know how to make the table to consider whether critical points are min/ max??

Answer 9

take steps, be patient.
now, tell me x^2-16 has 2 roots -4 and 4, then when (x^2 -16)>0? when it is<0 ??

Answer 10

you can check by let a value of x and calculate x^2 -16 to see my table is ok
for example, if x = 1, then x^2 -16 = -15 <0 , right? so I put negative sign between -4 and 4 to show you on that interval, x^2 -16 has negative value.

Answer 11

if you let x =5, then x^2 -16 = 25-16 =9 >0 , that is why the sign of x^2 -16 is + from the right of 4
the same with the left of -4
got it?

Answer 12

yes

Answer 13

now, consider x =2, when x =2, x -2 =0, so that
got it?

Answer 14

yes

Answer 15

now I copy it to the first one

Answer 16

got it?

Answer 17

not yet.

Answer 18

now pay attention at this
if I take the first row / the second row, you have g' , right?