Question

how did 3x+2y-7=0 became the equation of the tangent line if the given is f(x)= 2/fourthroot(x^3) and P(1,2) please explain it to me :(

Answer 1

\[\Large f(x)=\frac{ 2 }{ \sqrt[4]{x^3} }= 2x^{-3/4}\]

\[\Large f\prime(x)= -\frac{ 3 }{ 2 } x^{-7/4}\]

Did you get that much so far?

Answer 2

If so, then evaluate \(\Large f\prime(1)\). That gives you the slope of the tangent line.

You are given a point on the line.

Plug the slope and point into whatever form of line equation suits your fancy (I used point-slope) and solve so that you have the form Ax + By + C=0

You should get the line given. :)

Answer 3

i got the first one.. how did the second one became -3/2? i do understand the happening in the exponent though.. im sorry im not good in math :(

Answer 4

Do you know how to take a derivative? (I'm assuming so, in light of this question.)

Answer 5

yes i do, but i'm not really good at it.. i can get derivatives of simple equations, but when the given becomes complicated, i am lost of finding its derivative :(

Answer 6

\[\Large \left[ Cx^n \right]\prime=Cnx^{n-1}\]

Answer 7

Ok, NO NEED FOR SAD FACE. :) Math is happy. We'll just go through it.... hang on....

Answer 8

\(\Large f(x)= 2x^{-3/4}\)

\(\Large f\prime(x)= 2\cdot \dfrac{-3}{4} x^{-3/4-1}\)

Answer 9

\[\Large \cancel2\cdot \dfrac{-3}{\cancel4^2} x^{-3/4-4/4}\]

Answer 10

got that :)

Answer 11

\(=\Large \dfrac{-3}{2} x^{-7/4}\)

Answer 12

K? Good on that? That's your derivative. :)

Answer 13

yes.. thank you so much :) big big help for me.. :) :) :) :) :)

Answer 14

I did it prettier the first time.... lol

\(\Large f\prime(x)= -\frac{ 3 }{ 2 } x^{-7/4}\)

Answer 15

Great! You ok from here?

Answer 16

i hope so! i'm currently trying to solve it.. :) but thank you, youre a big help. i understood it now :)

Answer 17

Happy to help! :)

Answer 18

3x+2y-7=0 first solve for y

you'd get

y=(-3 /2)x+7/2
---
Maybe you dont have to but to me it is easier with y=mx+b form

2/(x^3) ^(1/4) BTW ^(1/4) is like saying 4th rooth

I wanna try it graphically

https://www.google.com/search?q=(-3+%2F2)x%2B7%2F2%2C+2%2F(x%5E3)+%5E(1%2F4)&oq=(-3+%2F2)x%2B7%2F2%2C+2%2F(x%5E3)+%5E(1%2F4)&aqs=chrome..69i57.5801j0&sourceid=chrome&ie=UTF-8

As you can see it is a tangent :)

Answer 19

wow, thank you for that :) i'm a fan :)))) got it! thanks for the graph, big help @timo86m :)

Answer 20

Here is another cool graphing tool.... I like that you can put the equation in as-is, don't have to solve for y. :)

It's always fun to "see" the math!! It doesn't help you DO the problem, but it does help you see the connection visually, and have confidence that the answer is right

http://kevinmehall.net/p/equationexplorer/#3x+2y-7=0 |2/%28x^%283/4%29%29|[-10,10,-10,10]

Answer 21

algebraically tho i think you would need to solve a system of equations

2/(x^3) ^(1/4)=y
(-3 /2)x+7/2 = y which is

(-3 /2)x+7/2= 2/(x^3) ^(1/4) I think that is how it is done. the solution should be
P(1,2)

Answer 22

ugh, you'll have to c&p that broken link.

Answer 23

http://www.wolframalpha.com/ works too :)

input more than 1 equation or expressions with a ,

Answer 24

algebraically tho i think you would need to solve a system of equations

2/(x^3) ^(1/4)=y
1 2
(-3 /2)x+7/2 = y which is
1 2


(-3 /2)x+7/2= 2/(x^3) ^(1/4) I think that is how it is done. the solution should be
P(1,2)
---p 1,2

Answer 25

thank you to both of you, i really appreciate it :)

Answer 26

Also dont forget
f'(x) gives you the slope at x. It doesn't give you an equation to draw it for you.

y=mx+b draws a line for you. But f'(x) only gives you the m part of it.

so lets say f'(3)=2 then you know the slope aka m at c=2

y=2*3+b plug the known info to y=mx+b: m and x


y is gotten thru f(3)= y lets say it is 4 f(3)= 4



4=2*3+b now you simply solve for b cuzz you got enough info for it.
b=-2 now you can you plug x and y back to the traditional y=mx+b

y=2x-2

Answer 27

thank you so much @timo86m wow, math wizard :)

Answer 28

this will be how it looks graphicaly

Answer 29

that is just the tangent line but f(x) could look like many things






Anything really so long as it is a tangent at 3,4