Question

Let g be a function that is defined for all x, x≠2, such that g(3)=4 and the derivative of g is g′(x)=
x^2–16/x-2, with x ≠ 2.

1. Find all values of x where the graph of g has a critical value.
I found that the critical values are x=4 and x=-4.

2. For each critical value, state whether the graph of g has a local maximum, local minimum or neither. You must justify your answers.
3. On what intervals is the graph of g concave down? Justify your answer.
4. Write an equation for the tangent line to the graph of g at the point where x=3.

Answer 1

@freckles

Answer 2

@jim_thompson5910 @SolomonZelman

Answer 3

will give medal ! (:

Answer 4

Critical values, slope of the derivative is zero, or non existent

Answer 5

Then make a number line with your critical values, and test if it is increasing or decreasing on either side of those values

Answer 6

Take the derivative of g'(x) to get g''(x)

Set g''(x) to zero to find inflection points, test those the same way to find concavity

Answer 7

evaluate the first derivative at x=3, that is the slope of the tangent line, at the point(3,g(3)) = (3,4)

Answer 8

use points slope form for the equation of the tangent line... slope g'(3) and point (3,4)

Answer 9

thank you! this is a great explanation, can you write out all the steps so i can understand it better? :)

Answer 10

Ok, let us walk through it.

Answer 11

okay so to start off, are the critical values of x=4 & x=-4 correct?

Answer 12

Is this the correct g'(x), you didnt use parenthesis , so i am guessing from your crit #'s

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

Answer 13

yes! perfect :)

Answer 14

ok, i integrated that , and graphed it

Answer 15

just to see what it looks like

Answer 16

anyways, you set that to zero and got, +4 and -4 as crit points

Answer 17

yup!

Answer 18

now i have to determine the nature of the graph for each crit point

Answer 19

so do i just plug in -4,2,&4 into g'(x)=x^2-16/x-2?

Answer 20

i would test first g'(-5), then g'(0), then g'(3), then g'(5)

Answer 21

ohh okay

Answer 22

no it has to be inside the interval, not the actual crit points

Answer 23

how do you determine the interval

Answer 24

the critical points, and undefined points, like i drew on the number line above.
(-infinity, -4) (-4,2) , (2,4), and (4, infinity)

Answer 25

pick numbers somewhere inside those intervals, and test if g ' is + or - value

Answer 26

For example

g'(-5) = - 9/7 the graph is decreasing on the interval (-infinity, -4)

Answer 27

okay so g'(-5)=-9/7 negative value
g'(0)= 8 positive
g'(3)=-7 negative
g'(5)=3 positive

Answer 28

So you get that chart

Answer 29

yes (:

Answer 30

x = 2 is stated as undefined, you can draw a vertical dotted line there, vertical asmyptope thing

Answer 31

and on either side of 2, it is increasing and decreasing, so it will look something like