Equation for Tangent Line to Inverse Function at (3,1) of f(x)=x^3+2x^2-x+1

Click For Summary
SUMMARY

The discussion focuses on finding the equation of the tangent line to the inverse function of f(x) = x³ + 2x² - x + 1 at the point (3, 1). The Inverse Function Theorem is applied, yielding the derivative 1/(3x² + 4x). By substituting x = 3, the slope of the tangent line is determined to be 1/21. The final equation of the tangent line is expressed as y = (1/21)x + 6/7.

PREREQUISITES
  • Understanding of the Inverse Function Theorem
  • Knowledge of derivatives and slope calculations
  • Familiarity with polynomial functions and their properties
  • Ability to manipulate linear equations
NEXT STEPS
  • Study the Inverse Function Theorem in detail
  • Practice finding derivatives of polynomial functions
  • Learn how to derive equations of tangent lines
  • Explore the relationship between a function and its inverse
USEFUL FOR

Students studying calculus, particularly those focusing on inverse functions and tangent line calculations, as well as educators seeking to clarify these concepts.

kathrynag
Messages
595
Reaction score
0

Homework Statement


Find an equation for the line tangent to to the graph of f^-1 at the pt (3,1) if f(x)=x^3+2x^2-x+1



Homework Equations





The Attempt at a Solution


I used the Inverse Function Thm
1/(3x^2+4x)
Now do I plug in 3 to this to find slope?
1/21
y-1=1/21(x-3)
y-1=1/21x-1/7
y=1/21x+6/7
 
Physics news on Phys.org
state the inverse function thm.
what is (3,1) a point of?

also, i think you are mixing up x's. if you look at f^-1 as a function of x, these x's are really values (or y's) of f(x)

i hope you understand this; I'm afraid I've been a bit convoluted.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
Replies
13
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
Replies
2
Views
1K
Replies
5
Views
1K