Question

Let g be a function that is defined for all x, x≠2, such that g(3) n=4 and the derivative of g is g′(x)=
x2–16
x−2
, with x ≠ 2.
Find all values of x where the graph of g has a critical value.
For each critical value, state whether the graph of g has a local maximum, local minimum or neither. U must justify your answers with a complete sentence.
On what intervals is the graph of g concave down? Justify your answer.
Write an equation for the tangent line to the graph of g at the point where x = 3.
Does this tangent line lie above or below the graph at this point? Justify your answer

Answer 1

@eliassaab Please helpp!!! will give out medal :)

Answer 2

@undeadknight26 Please helpp!!! will give out medal :)

Answer 3

such that g(3) n=4 what does this mean?

Answer 4

is this what you mean (x2–16) /(x−2)? so that x =4, -4

Answer 5

g' would be the slope when the slope = 0 then x^2 - 16 = 0 to give those values of x
already given that x not = 2 (we will not divide by zero)

Answer 6

one way to check for maximum or minimum is to get the 2nd derivative at that critical point positive means minimum and negative for maximum

Answer 7

you can apply the quotient rule to get the derivative

Answer 8

Write an equation for the tangent line to the graph of g at the point where x = 3.
the slope of the tangent line = g'
when x = 3 ; replace x with 3 in the equation and compute
y - g(3) = -7(x - 3)
g(3) n=4 if the n is an error and g(3) = 4 then
y - 4= -7(x-3)
y= -7x + 21 + 4
tangent line at x = 3
y = -7x + 25

Answer 9

Yes i'm so sorry I meant g(3)=4 the n is an error @triciaal

Answer 10

so can you please explain it from the beginning i'm a little confused because of the n that i put (:

Answer 11

so how do i find all values of x where the graph of g has a critical value? :D @triciaal

Answer 12

please review all my entries.
I assumed it was an error later.
I went through the whole problem!
what specific question do you have?

Answer 13

@triciaal, you method makes sense to me...

Answer 14

great. thanks.

Answer 15

awesome! i understood the explanation :) thanks so much i just had to read it over a couple of times
doers this tangent line lie above or below the graph at this point? :) Justify the answer

Answer 16

@triciaal

Answer 17

i know i have local minimums for both 4 and -4, and an asymptote at x=2.
my intervals for the graph are: to the left--> -infinity<x<2 and to the right---> 2>x>infinity

Answer 18

I think all i have to do is plug in values for x and see if it goes above or below the graph, correct ? :)

Answer 19

if taht's the case then i think it lies above

Answer 20

you have a zero at x = 4

Answer 21

no u have a -3
-7(4)+25
-28+25 = -3

Answer 22

@SolomonZelman

Answer 23

@Directrix

Answer 24

ohh its below

Answer 25

i think im not sure :P

Answer 26

did you graph the functions?