Question

f(x)=5x((5-5x)^(1/2))

find the equation of line tangent to the graph f at x=-2

Answer 1

so have you found f'(x)?

Answer 2

don't think it's the right one

Answer 3

\[f'(x)=5[1*(5-5x)^\frac{1}{2}+x*\frac{1}{2}(5-5x)^{\frac{1}{2}-1}(-5)]\]

Answer 4

\[f'(x)=5*\sqrt{5-5x}-\frac{5}{2\sqrt{5-5x}}=\frac{5\sqrt{5-5x}*2\sqrt{5-5x}-5}{2\sqrt{5-5x}}\]
=\[\frac{10(5-5x)-5}{2\sqrt{5-5x}}=\frac{50-50x-5}{2\sqrt{5-5x}}\]
=\[\frac{-5x+45}{2\sqrt{5-5x}}*\frac{\sqrt{5-5x}}{\sqrt{5-5x}}=\frac{(-5x+45)\sqrt{5-5x}}{2(5-5x)}\]
=\[\frac{-5(x-9)\sqrt{5-5x}}{2(5)(1-1x)}=\frac{-(x-9)\sqrt{5-5x}}{2(1-x)}\]

Answer 5

i was just trying to make it look nicer

Answer 6

anyways to find the slope find f'(-2)

Answer 7

so I plug it in an my x value at -2 is 7.1005. I'm stuck after that.

Answer 8

\[f'(-2)=\frac{-(-2-9)\sqrt{5-5(-2)}}{2(1-(-2))}=\frac{11\sqrt{5+10}}{6}=\frac{11\sqrt{15}}{6}\]

Answer 9

do we know a point in the tangent line at x=-2?
yes we (-2,f(-2)) on on that line
\[f(-2)=5(-2)\sqrt{5-5(-2)}=-10\sqrt{15}\]
a line has the form y=mx+b
m=11sqrt{15}/6
(x1,y1)=(-2,-10sqrt{15})
\[-10\sqrt{15}=\frac{11\sqrt{15}}{6}*(-2)+b\]
find b to find the y-intercept of the tangent line

Answer 10

\[b=-10\sqrt{15}+\frac{11\sqrt{15}*2}{6}\]
\[b=\sqrt{15}(-10+\frac{11*2}{6})=\sqrt{15}(-10+\frac{11}{3})=\sqrt{15}(\frac{-30+11}{3})\]
\[b=\frac{-19}{3}\sqrt{15}\]

Answer 11

I get 7.100469468x+52.9307724, which is wrong. What am I doing?

Answer 12

\[y=\frac{11\sqrt{15}}{6}*x+\frac{-19\sqrt{15}}{3}\]

Answer 13

first does say to find the exact line or an appoximation
also are y-intercepts are way different

Answer 14

all it says is find the equation of line tangent to the graph, and tangent line : y=?

Answer 15

I plugged it in, still not getting it

Answer 16

line has form y=mx+b
we know a point on the line and we know the slope so we can find the y-intercept

Answer 17

what did you plug in for the point?

Answer 18

exactly what you gave me, and others that haven't worked either.

Answer 19

how did you get a different y-intercept then?

Answer 20

incorrect math, I guess.

Answer 21

Good work myininaya. These computer exercise would reject an answer for little mistakes. There is a little mistake at the derivative. Find derivative of inner function (5-5x); the derivative is -5

Answer 22

yeah i have that above

Answer 23

I give up.

Answer 24

show me what you did when you tried to find the y-intercept

Answer 25

you said you plug in exactly what i gave right and got different y-intecept than me right?
y=mx+b
b=y-mx
b=f(-2)-f'(2)(-2)

Answer 26

b=f(-2)+2f'(2)

Answer 27

b=-10sqrt{15}+2(11sqrt{15}/6)

Answer 28

you should get a negative y-intercept since 10>11/3

Answer 29

b=sqrt{15}*(-10+11/3)
b=sqrt{15}*(-30/3+11/3)
b=sqrt{15}*(-19/3)

Answer 30

Try this
y=25.8207x + 12.9099

Answer 31

chaguanas, that worked :] however did you find it?

Answer 32

i think i forgot to distrubute the 5 earlier when i was doing f'(x)

Answer 33

I used the KISS method: Keep it simple. :)

Answer 34

I had no idea what I was doing wrong. Do you think you could help me with thesame type of problem? f(x)= (3x^3)/((4-5x)^5), equation of line tangent to graph of f at x=2.

Answer 35

Myininaya, you were doing it right, but this is one of those computerized thingies. You made a lot of transpositions and transformations to 'make it look pretty' and you might have threw it off one percentage point. Let's see this one.

Answer 36

i did notice some errors above

Answer 37

i forgot to distribute something
and then forgot to add something lol

Answer 38

\[f'(x)=[9x ^{2}(4-5x)^{5}+75x ^{3}(4-5x)^{4}]/(4-5x)^{10}\]

Answer 39

I get \[f \prime(x)= 6x ^{2}(5x+6)/(4-5x)^6\]

Answer 40

with that, I get m=.00823

Answer 41

y=mx+b
-1/324=(2/243)*2+b
b=-1/324-4/243
b=-19/972
y=2/243*x-19/972
but maybe i made a mistake again lol

Answer 42

Show us how you got that f'(x)

Answer 43

your f' looks delicious gj

Answer 44

did you look at the attachment yet?

Answer 45

That was for mnad1, looks like he made mistakes in derivative. I have an old computer and I can't read a lot of the attachments.

Answer 46

your answer is correct, myininaya. Thank you! I'm annoyed because I get a completely different derivative. Looking at your attachment I see that I deviate from the correct steps pretty much at the beginning

Answer 47

but you guys are awesome, thanks for the elaborate explanations, it really helped

Answer 48

you got 6x^2(5x+6))/(4-5x)^6 right?

Answer 49

ok well your welcome lol

Answer 50

OK, I keep telling him that is wrong. How did you get that myininaya, compare it with my derivative about five messages up.

Answer 51

yours is right too, i just factored (4-5x)^4 out on top and canceled (4-5x)^4 on top and bottom leaving me with
on top: 9x^2(4-5x)+75x^3
on bottom: (4-5x)^6
we have like terms on top:36x^2-45x^3+75x^3=36x^2+30x^3=6x^2(6+5x)

Answer 52

yeah, that was my f'(x). His was the same as yours, and you were evidently both right.

Answer 53

hers*

Answer 54

i was just going for pretty looks again lol

Answer 55

OK, mnad1, sorry I keep saying you're wrong. But in a case like this you are inputting values of x, so there is no need to break it down further. It increases the chance of making a mistake. Only a pro like myininaya could do that.

Answer 56

lol

Answer 57

but as we learned i can make mistakes too example the first one we did

Answer 58

yes simplifying things can be lead to problems like mistakes so if your teacher doesn't care about simplying (or making pretty) then just do what chag says well unless you just enjoy it like i do

Answer 59

but mnad the form we have it in
we can easily find horiztonal tangents :)

Answer 60

I guess you could say I'm half assing it cause in order to get the x value, I just plug the derivative in my calc and look at the table. It's not a problem unless the derivative is wrong, as it was in my case and I had no idea what my error was. until now.

Answer 61

we can say we have a horizontal tangent at x=0 and x=-6/5
if we were asked to find this

Answer 62

I don't know if you noticed, you might have, but math is not my best class alas.

Answer 63

i didn't notice

Answer 64

be positive and don't give up
if you believe you can't do, then chances are you probably won't be able to do it.
so just believe you can

Answer 65

it's a summer class so we're going through the material at a faster pace, so I have to apply extra effort to learn it. I'm glad I came across OS though, the people here are extremely helpful.