MHB Katie's Question: Find f(2) & f'(2) of Tangent Line y=4x-5

Click For Summary
SUMMARY

The discussion focuses on finding the values of f(2) and f'(2) for the tangent line equation y=4x-5 at the point where a=2. The function under consideration is f(x)=x^2-1. The derivative f'(2) is determined to be 4, matching the slope of the tangent line, while the value of the function at that point, f(2), is calculated to be 3, which corresponds to the y-value of the tangent line at x=2.

PREREQUISITES
  • Understanding of calculus concepts such as derivatives and tangent lines.
  • Familiarity with function notation and evaluation.
  • Knowledge of basic algebraic manipulation.
  • Ability to interpret graphical representations of functions and their tangents.
NEXT STEPS
  • Study the concept of derivatives in calculus, focusing on how to find the slope of a tangent line.
  • Learn about the relationship between a function and its tangent line at a given point.
  • Explore the graphical interpretation of functions and their derivatives using tools like Desmos or GeoGebra.
  • Practice evaluating functions and their derivatives at specific points with additional examples.
USEFUL FOR

Students learning calculus, educators teaching derivative concepts, and anyone interested in understanding the relationship between functions and their tangent lines.

MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

If an equation of the tangent line to the curve y=f(x) at the point where a=2 is...?


y=4x-5, find f(2) and f'(2)

Please walk me through this question step by step.

I have posted a link there to this topic so the OP can see my work.
 
Physics news on Phys.org
Hello Katie,

Consider the following diagram, showing the function $f(x)=x^2-1$ and its tangent line $y=4x-5$:

View attachment 1403

As you can see, the instantaneous slope of the function must match the slope of the tangent line at the point of tangency, and the value of the function at that point must be equal to the value of the tangent line at that point since they touch there. Hence, we must have:

$$f'(2)=\frac{d}{dx}(4x-5)=4$$

$$f(2)=y(2)=4(2)-5=3$$
 

Attachments

  • katie.jpg
    katie.jpg
    7 KB · Views: 124
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

Similar threads

Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
Replies
5
Views
2K