• Support PF! Buy your school textbooks, materials and every day products Here!

Tangent of a line

  • Thread starter menco
  • Start date
  • #1
43
0

Homework Statement


The graph of y = f(x) passes through the point (9/2, 100/3). Also the tangent line to the graph at any point (x,y) has the slope 4*sqrt(2x+7). Find f(x)


Homework Equations





The Attempt at a Solution


I am very lost with this as I can't find much info in my text book. Any help where to start? I am assuming I am trying to find the equation of the line y=mx+c given the slope of the tangent and current points.
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6


Homework Statement


The graph of y = f(x) passes through the point (9/2, 100/3). Also the tangent line to the graph at any point (x,y) has the slope 4*sqrt(2x+7). Find f(x)


Homework Equations





The Attempt at a Solution


I am very lost with this as I can't find much info in my text book. Any help where to start?
Given a function [itex]y=f(x)[/itex], how does one find the slope of the tangent line? (If you aren't sure, you had better open up your textbook and find out the definition of tangent line)

I am assuming I am trying to find the equation of the line y=mx+c given the slope of the tangent and current points.
No, you are asked to find the original function [itex]f(x)[/itex] (which will not be a straight line), not the tangent line at some point.
 
  • #3
43
0


I know how to find the slope and equation of a tangent line fairly easy but trying to reverse it is confusing me
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6


I know how to find the slope and equation of a tangent line fairly easy
Again, describe how to find the slope of the tangent line to a function [itex]y=f(x)[/itex]. (Don't say that you know how, demonstrate that you know)
 
  • #5
43
0


To find the slope of a tangent line take the derivative of the function and substitute in the point of contact if known.
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
6


To find the slope of a tangent line take the derivative of the function and substitute in the point of contact if known.
Right, and so at a general point [itex]x[/itex], the slope is just [itex]f'(x)[/itex]. So, what can you say about [itex]f(x)[/itex] if the slope of the tangent line at a point [itex]x[/itex] is [itex]4\sqrt{2x + 7}[/itex]?
 
Last edited:
  • #7
43
0


the rate of change of the function is 4*sqrt(2x+7)?
 
  • #8
gabbagabbahey
Homework Helper
Gold Member
5,002
6


the rate of change of the function is 4*sqrt(2x+7)?
Yes, [itex]f'(x)=4\sqrt{2x+7}[/itex].

So, [itex]f(x)=?[/itex]...
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,833
955


The opposite of the derivative is the anti-derivative, also called the "indefinite integral". Have you studied those?
 
Last edited by a moderator:
  • #10
43
0


Yes we have just started integrals, by using substitution

I found f(x) = (4(2x+7)^3/2) / 3

Does the point (9/2, 100/3) have anything to do with the problem?
 
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257


Yes, the graph of y=f(x) passes through that point, so you should have that 100/3 = f(9/2). Is that the case for your f(x)? If not, how do you fix it?
 
  • #12
43
0
f(9/2) = 256/3

So it is not the case, I'm a little unsure of what you mean by fix it?
 
  • #13
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
##\frac{4}{3}(2x+7)^{3/2}## is not the only function whose derivative is ##4\sqrt{2x+7}##. You need to find another one, one where f(9/2)=100/3.
 
  • #14
5
0
menco, take this equation for example: [itex]\int x^2 = \frac{x^3}{3} + C[/itex], do you remember why we put +C there?
 
  • #15
43
0
Ah yes i see I forgot all about +C, which is the constant of integration.

So if I use 100/3 = 256/3 + C, C = (-52)

Therefore the final function will be ((4(2x+7)^3/2) / (3)) - 52

then when f(9/2) = 100/3
 

Related Threads on Tangent of a line

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
378
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
591
  • Last Post
Replies
7
Views
3K
Top