Question

I am on the verge of a panic attack, i missed 3 weeks of school because i was sick. I am still sick but im not throwing up anymore so i am attending school. I hae finals tomorrow and i need everything to be able to get ready! I have a lot of questions and not much time, so it would be the quickest way if people could answer and explain. I am going to be up all night and i need help!!!!!!

Answer 1

Post some stuff!

Answer 2

for \[f(x) = \frac{ x ^{2} }{ x-1 }\] use the definition of differentiability \[f \prime (c) \frac{ f( x +\Delta x) - f(x) }{ \Delta x }\] to determine if f(x) is differentiable at x=1

Answer 3

@stardogchampion

Answer 4

hi wat grade are u in

Answer 5

12th

Answer 6

wat school

Answer 7

im taking finals 2

Answer 8

hello

Answer 9

any1 here

Answer 10

it can't be differentiable at 1 because the function is not even defined at 1

Answer 11

this is clear right? if the function does not exist somewhere, obviously it's derivative doesn't

Answer 12

i think she left

Answer 13

yes it is thank you! do you have time to help me more???

Answer 14

sure

Answer 15

btw we did not bother to actually compute \(f \prime (c)= \frac{ f( x +\Delta x) - f(x) }{ \Delta x }\) because it is a stupid idea, but we can try it if you like and see why it does not exist

Answer 16

i don't see the point in that if we are going to get the same answer. can you explain why functions with corners are not differentiable even though it is continuous?

Answer 17

which in any case should be \[f \prime (x) =\frac{ f( x +\Delta x) - f(x) }{ \Delta x }\]or
\[f \prime (c) =\frac{ f( c +\Delta x) - f(c) }{ \Delta x }\]

Answer 18

yes i think so

Answer 19

intuitively it is because the derivative evaluated at a number is the slope of the line tangent to the graph at that number
if the function has a corner at that number, there is no definition for one tangent line
as in this picture

Answer 20

reason two
the derivative is a limit
in order for a limit to exist it has to exist from the left, and from the right, and those limits have to be equal
in the famous example of why \(f(x)=|x|\) does not have a derivative at the corner \((0,0)\) is because the limit of the difference quotient from the left is \(-1\) and from the right is \(1\) and those number are not the same

Answer 21

okay thank you.

Answer 22

yw

Answer 23

Use the following graph of f(x) to describe the behavior of f ‘(x). State where f ‘(x) is positive, negative, zero or not defined.

Answer 24

is it positive? or not defined?

Answer 25

it is not defined where the function is not defined, at \(x=2\)

Answer 26

so any with an asymptote is not defined?

Answer 27

the derivative is the slope of the tangent lines
the slopes will be positive over the intervals where the function is increasing, i.e. going up

Answer 28

function doesn't exist at the vertical asymptote, neither does the derivative

Answer 29

derivative will be negative over the interval where the function is decreasing

Answer 30

okay, i get that

Answer 31

looks like your function is increasing over \((-3,2)\cup (3,5)\) and decreasing on \((2,3)\)

Answer 32

derivative will be zero where the tangent line is horizontal, if i am reading the graph correctly looks like it will be horizontal as \(x=3\)

Answer 33

a good candidate for a local maximum or a local minimum
in this case \(x=3
\) looks to be a local min

Answer 34

Use the following graph of f (x) to find the graph of f ‘(x).

Answer 35

that function is decreasing, then increasing
look for a derivative that is negative (below the x axis) then positive (above the x axis)

Answer 36

i wanna say it is the last one?

Answer 37

that is a tough one
i might say 143
if you look at the rate of decrease of the original function, over one interval it looks like it is decreasing at a constant rate
if you look at the derivative 143, you see it is almost horizontal over the same interval

Answer 38

gotta run, back later
post again if you have more questions in a new thread

Answer 39

kk thank you!