Question

For what value a is the tangent line to the graph f(x) = -x^2 + x +c at the point (a,f(a)) parallel to the line y= -x+3

Answer 1

you have a tangent line (whose slope is determined by the derivative) parallel to the given line y=-x+3 whose slope is...

Answer 2

i found the derivative of -x^2 + x + which is -2x + 1

Answer 3

and i equate -2x + 1 = -x +3

Answer 4

and is wrong ^^ lol

Answer 5

for some x, the slope of the tangent line is equal to the slope of the line y = -x + 3 (whose slope is -1)

Answer 6

solve x in -2x + 1 (slope of tangent line) = -1 (slope of the given line)

Answer 7

but the given answer says +1

Answer 8

oh yes

Answer 9

slope is -1 , a = 1

Answer 10

but what's the point of -x^2+x+c

Answer 11

the is the curve where one of its tangent lines (whose slope is determined by its derivative) is parallel to some given line, in this problem, y= -x + 3.
if there's no curve, there's no point discussing tangent lines.

Answer 12

well, there is no graph given
so basically if the function f(x) has a tangent whose equation is given, the slope AT ONE POINT of the function is the same as y=-x+3

where as -2x+x defines the slope at all points

then can't I equate -2x+x = -x+3 to find the point that they are equal?

Answer 13

-2x + 1 is THE slop of tangent line, while -x + 3 is a line. you can't equate the slope and a line. it must be slope = slope.

Answer 14

then slope of the line -x+3 = -1
-2x+1 = -1
x= 1

oh.. lol

Answer 15

gotcha

Answer 16

oh thanks!!
and then another one
find derivative of x^cos(x)

i just forgot the formula to do that..

Answer 17

let y = x^cos x
take the logarithm first of both sides, then use implicit differentiation

Answer 18

can i use product rule

Answer 19

i first did it and got -sinxlnx + cosx / x
but i think the answer should include xcos^x in it

Answer 20

yes, after taking the logarithm of both sides.

Answer 21

still dont get it lol

Answer 22

after u log it , it becomes cos x ln x..
then where is the x ^cosx

Answer 23

where u go @@

Answer 24

y=x^cos x
ln y = ln (x^cos x)
ln y = cos x ln x

now differentiate both sides
(1/y) dy/dx = cos x (1/x) - sin x ln x
solve dy/dx

Answer 25

oh ok thanks!
and when i ask where u go, basically just to make sure u received the msg, so is not u wait for me , and i 'm waiting for u as well.
so if u have anything to do, u can always just come back after

Answer 26

im having problems with my net connection.

Answer 27

oh ok, and np, i will just wait :P

Answer 28

and then #17, just sent u up there

Answer 29

\[\Large f'(x)=\begin{cases}1, & x>2\\-1, & x<2 \end{cases}\]

Answer 30

notice that in the derivative, the "=" disappeared in \(x \leq 2\)

Answer 31

yes, that means it is not differentiable at x=2

Answer 32

right. differentiability is assured only on an open interval, not at an endpoint. in this case, x=2 is an endpoint.

Answer 33

oh, answer is E
not A because it includes x=2 being differentiable
not B because of the same reason
not C because x=2 is not differentiable
Not D because everywhere is differentiable except x=2
so is E!

Answer 34

and what does the 1, and -1 mean < i know is the slope, but what kind of info does it give?

Answer 35

f'(x) =1 , x > 2 is telling us that the derivative exists and its value is 1

Answer 36

ohh oh yea, thanks!
if u need to do my test, i feel like u are gonna finish it in 15mins ..>.>

Answer 37

and #20 plz :D

Answer 38

what do you think?

Answer 39

It was on my midterm and I just gave up lol

Answer 40

for the dot product, i believe they need to be differentiable
and i think there is a rule for that

Answer 41

but I dont remember the rule, and no idea where to find it..and plus those options are confusing enough

Answer 42

it's (D).
THe derivative of \(f \cdot g\) is \(f' \dot g + f \cdot g'\), PROVIDED \(f\) and \(g\) are differentiable.
the required premises for computing the derivative is the differentiability of the factors.

Answer 43

if a variable has a derivative, does it mean it is differentiable for sure

Answer 44

differentiable only on an open interval, NOT the closed interval.

Answer 45

i actually dont understand what D is saying @@

Answer 46

what is those "not guarantee" stuff... talking about open and closed intevals?

Answer 47

it says if f, g, f' and g' exists, then (1) (fg)' exists; and (2) we can use the formul (fg)' = f'g + fg'

Answer 48

but it also says differentiability of f exist doesnt mean g exist
but if f dot g exist, shouldnt g be differentiable as wlel

Answer 49

like..how can u put something differnetiable with something that is not differentiable and get something differentiable @@

Answer 50

it only says you know tha (fg)' exists. you cannot use the formula (fg)' = fg' + gf' since you are not assured that both g' and f' exists.

Answer 51

to translate it, if you know that material A is magnetically attracted to B, can you be certain that A is a magnet?

Answer 52

noo

Answer 53

ahhh...much more clear :P

Answer 54

thks for the analogy haha

Answer 55

(D) says how to compute (fg)' if you know f' and g'.

Answer 56

yea , i get it when u explain.
but I'm really weak in these theory/concept questions :/

Answer 57

basically if a ques has too many words , i will just be lost..

Answer 58

for x^2 - 2x^2 + 2x/ x^4
to find the derivative, is it better to break it in parts of just directly the difference rule

Answer 59

agree.

Answer 60

so break it by parts?

Answer 61

and then
for f(x) = x^1/3 , why Newtons' method will fail with x1 = 1
is it because the point it taken near the end point, which is the absolute minimum?

As I see Newton's method will fail if

1.) f'(x)=0
2.) The initial value is taken at a maximum or minumum point,
3.) The initial value is taken at a point of inflection,
4.) The initial value is taken near a maximum point and a minimum point,
5.) The initial value is taken near a point of inflection.

Answer 62

still thinking...

Answer 63

oh ok