Help with inverse of derivative function

  • Thread starter Hazy001
  • Start date
  • #1
2
0
Member warned about not using the homework template
f(x) =
sqrt2a.gif
3x^3 + 3x^2+ 2x + 1
,a = 3

formal is
img5.gif


Homework is due tonight and this is the only problem i cant solve

Your suppose to
3=
sqrt2a.gif
3x^3 + 3x^2 + 2x + 1
, solve for xThe find the derivative of y=
sqrt2a.gif
3x^3 + 3x^2 + 2x + 1
, then plug x into that and put it under 1.
 

Answers and Replies

  • #2
34,697
6,399
f(x) =
sqrt2a.gif
3x^3 + 3x^2+ 2x + 1
,a = 3
What does a represent? I infer from your work below that you are taking it to mean the point (x0, a) on the graph of f.
Hazy001 said:
formal is
img5.gif


Homework is due tonight and this is the only problem i cant solve

Your suppose to
3=
sqrt2a.gif
3x^3 + 3x^2 + 2x + 1
, solve for xThe find the derivative of y=
sqrt2a.gif
3x^3 + 3x^2 + 2x + 1
, then plug x into that and put it under 1.
Assuming that your interpretation is the correct one, start with this equation:
##3 = \sqrt{3x^3 + 3x^2 + 2x + 1}##, and then square both sides. The resulting equation is not the easiest to solve, but it does have one real solution.
 
Last edited:
  • #3
2
0
Yes that is it but sadly i cannot solve it
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
Yes that is it but sadly i cannot solve it
You want to solve ##p(x) = 0##, where ##p(x) = 3 x^3 + 3 x^2 + 2x - 8##.
(i)One thing to try is the "rational root theorem"; see, eg.,
http://en.wikipedia.org/wiki/Rational_root_theorem .
(ii) Alternatively, plot the graph of ##y = p(x)## over some ##x##-range, to see roughly where a root lies; that will suggest a factor of ##p(x)##, which you can then verify exactly. (iii) If you are still desperate you can always submit the problem to an on-line solver, such as Wolfram Alpha.
 

Related Threads on Help with inverse of derivative function

Replies
16
Views
5K
  • Last Post
Replies
7
Views
756
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
14
Views
789
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
11
Views
511
Replies
3
Views
2K
Top