Question

so I'm wondering how I can find the slope of the tangent line in this graph...

Answer 1

I need to find when g'(2) and f'(2). do I just guess?

Answer 2

ok I'm also wondering like when the line is going both directions... like when x=2 of the g(x) graph then which way do I look to see the slope of the tangent line? to the left or to the right? am I making sense? lol

Answer 3

Well for f(x) it's pretty straight forward.
That portion of the function (from x=-0.5 to x=3) is a straight line.
So the slope is constant.

If we can find the slope of that line segment, that value will also be the instantaneous slope anywhere inside that interval.

Answer 4

ah its you again!

Answer 5

haha :3

Answer 6

right i agree with you but I don't know how to find the slope of the tangent line without a function...

Answer 7

So I'm kind of estimating here, but it looks like we can grab a couple of coordinate points from f(x).
\[\large (3,3.5) \qquad \qquad (-0.5,2)\]

Using our formula for slope gives us,
\[\large m=\frac{f(3)-f(-0.5)}{3.5-2}\]

Answer 8

other than guessing?

Answer 9

ah :) ok

Answer 10

Ya you kinda have to guess a little bit it looks like D:

Answer 11

Woops I set that up really wrong :) one sec.

Answer 12

\[\large m=\frac{f(3)-f(-0.5)}{3-(-0.5)} \qquad\to\qquad m=\frac{3.5-2}{3-(-0.5)}\]
That makes more sense.. I had my y values in the denominator before. woops.

Answer 13

haha no worries.

Answer 14

1.5/3.5?

Answer 15

ya that sounds right :O maybe it would look better if we wrote it as a fraction.. i dunno.\[\large \frac{\left(\dfrac{3}{2}\right)}{\left(\dfrac{7}{2}\right)} \qquad=\qquad \frac{3}{7}\]

Answer 16

ok so that's my slope right?

Answer 17

Yah f'(2)=3/7.
We have to kind of hope that we chose our coordinate pairs well though :p

Answer 18

That's the slope of your line segment.
But it's also the instantaneous rate of change at any point in that interval :)
That's why that slope value works for our derivative.

Blah that's a weird way to say that :P hope that's not confusing.

Answer 19

About the question you had regarding g(x),

Answer 20

Assuming the function is continuous at x=2, we could look to the left OR right to determine the slope. Because if it's continuous, it will be a nice smooth connection at x=2 and the slopes will be the same from the left and from the right side.

If that was the case though (the function being continuous), we would simply choose to grab the slope from the left side because it's much easier to deal with a straight line as we saw with f(x).

Answer 21

If the function is not continuous then they probably wouldn't had asked us to find g'(2)... So I guess we're ok :O

Answer 22

ok~

Answer 23

So I think for g(x) we can do the same thing we did with f(x), hopefully :3

Answer 24

So what coordinate pairs are you coming up with?

Answer 25

so just double checking thats the slope of the tangent line right?

Answer 26

For f(x), we found the slope of a `secant line`. (A line passing through 2 points).
Since it's a straight line, any tangent line in that interval will have the same slope as the secant line we found.

So yes, we found the slope of the `tangent line` as well. :)

Answer 27

i have a few... but is it wiser not to use zero's?

Answer 28

ok :)

Answer 29

Mmm I don't think it matters, I would try to use the end points of that straight line though.

Answer 30

i estimated a couple points of (-1,-2.5) and (2,0)

Answer 31

I think that first coordinate pair is (-1,-1.5).
Check again real quick. The way the labeled the lines is really stupid. The number you read is the one above the line.

Answer 32

-2 is the bottom edge of the graph.

Answer 33

oops my bad XD

Answer 34

So toss that into your slope formula, what do you get? :3

Answer 35

-1.5/3

Answer 36

-0.5?

Answer 37

(-1, -1.5) (2,0)

\[\large m=\frac{f(2)-f(-1)}{2-(-1)}\qquad\to\qquad m=\frac{0-(-1.5)}{2-(-1)}=\frac{1}{2}\]I think it's positive 1/2 hmm.

Answer 38

i'm so sorry!!!!

Answer 39

yeah ur right!

Answer 40

So that's probably an acceptable value for g'(2).
Do you have an answer key or no?
I wish we could check our work :o

Answer 41

no answer key :( sadly but I was suppose to plug it in... to this product rule

Answer 42

(fg)'

Answer 43

so I did f'(2)*g(2)+f(2)*g'(2) am I on the right track?

Answer 44

oh btw they said x=2.

Answer 45

They want you to evaluate (fg)' at x=2? :o
Hmm looks like you're on the right track.

Answer 46

yeah

Answer 47

this sounds really dumb but for f'(2) I'm suppose to write in the slope of the tangent line or the derivative at x=2 right?

Answer 48

Correct :o

Answer 49

ok thanks alot!!! sorry I left for a couple minutes!!

Answer 50

np :3