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.

2. For each critical value, state whether the graph of g has a local maximum, local minimum or neither. You must justify your answers with a complete sentence.

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
5. Does this tangent line lie above or below the graph at this point? Justify your answer.

Answer 1

@triciaal i reposted the problem and fixed it:) sorry for the inconvenience! @ganeshie8

Answer 2

1. to find the critical values let the numerator of the 1st derivative equal zero and solve

so solve

\[0 = x^2 - 16\]

2. find the 2nd derivative.... and test the critical values to determine there nature.


3. Co the concavity do s simply sketch of the critical points.

4. you have the point (3, 4) you know the 1st derivative substitute x = 3 into the 2st derivative to find the slope of the tangent, then you can get the equation.

5. do a simple sketch.

hope it helps

Answer 3

@campbell_st do you mind reviewing what I did first?

Answer 4

Nope! (: @triciaal i already reviewed it I just think i would understand it better if we talked through the steps 2gether :)

Answer 5

oh

Answer 6

ok... so what critical values did you get...?

Answer 7

I'm not sure how to find the 2nd derivative

Answer 8

am i finding the 2nd derivative of the numerator? :D @campbell_st

Answer 9

ok... well I'll talk you through an alternative method... start with the critical values...?

Answer 10

just ignore the 2nd derivative for a moment

Answer 11

:) yeah let's start with that,
so in order to find all values of x where the graph of g has a critical value, what do i do? :D

Answer 12

exactly what I said. set the 1st derivative to zero and solve for x.

so

\[0 = \frac{x^2 -16}{x -2}\]

the denominator can't be zero so you are really looking at solving

\[0 = x^2 - 16\]

Answer 13

what are the critical values if you solve this?

Answer 14

i'm suppose to solve for x correct? if i solve for x i get: x=-4,4

Answer 15

so there's 2 critical values

Answer 16

or is it just 4

Answer 17

because the square root of 16 is 4

Answer 18

ok great... and you need to know that the 1st derivative is the equation of the slope at any point on the curve....

so g'(4) = 0 and g'(-4) = 0...

I hope that makes sense...

now if I plot them, you can choose points either side of the critical values and look at the slopes to determine the nature of the critical values.



can you find g'(-5) ?

Answer 19

i'm suppose to state whether the graph of g has a local max or min or neither

Answer 20

then g'(-3), g'(3) and lastly g'(5)

I chose the points either side as -5, -3 3 and 5

by finding the slope you can determine the nature

Answer 21

ohhh ok :)

Answer 22

how do i plot the points tho

Answer 23

g'(-5) = - 9/7 so the slope is negative

Answer 24

you can finish it to look at the critical values

Answer 25

g'(-3)= -7/-5
g'(3)= -7
g'(5) = 3

Answer 26

great so looking at it

so looking at that, the critical values are both both critical points are minimums