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

Beginner(self study) calculus question.

  • #1
445
0

Homework Statement



Find the slope of the tangent line to the parabola y = 2x - x2 at the point (2,0)

Homework Equations


[tex]\frac{f(x) -f(a)}{x - a}[/tex]

[tex]\frac{f(a + h) - f(a)}{h}[/tex]



The Attempt at a Solution



[tex] \frac{(2x - x^2 - (2(2) - (2)^2)}{ x - 2} [/tex]

[tex] \frac{(2x - x^2 - ( term self cancels )}{ x - 2} [/tex]

[tex] \frac{x(2 - x)}{x - 2} [/tex]

This is where I'm stopped. As I wrote in the thread title, I'm independently studying the subject before I have to take the course, so I don't have anyone to direct me in this endeavor. Any help would be appreciated.
 
Last edited:

Answers and Replies

  • #2
1,341
3
To find the slope you have to take the limit as x approaches a, by the definition of derivative.
 
  • #3
56
0
Here's one way to do it:

1. Use the definition of derivative to find the derivative function.

f'(x) = [tex]\stackrel{lim}{h\rightarrow0}[/tex] [tex]\frac{f(x+h)-f(x)}{h}[/tex]

2. At the point (2,0), x = 2, so plug in 2 for x in the derivative function.
 
  • #4
1,341
3
What Dunkle said works too, if you'd like to obtain a function that will tell you the slope at every point.

Of course the way to were doing it works too, you just have to take a limit as x approaches 2 in your case.
 
  • #5
ZJM
1
0
The problem is that the difference quotient is a function in a (or h). So think of your first equation as a template. If you want to figure out the slope of the tangent line to (2, 0), plug in 2 for x in the difference quotient to obtain:

[tex]f\prime(2) = \lim_{a\rightarrow 2}\frac{f(2) - f(a)}{2 - a}[/tex]

What this equation is telling you is that the derivative of f at 2, i.e., the slope of the tangent at (2, 0), is equal to the limit of [tex]\frac{f(2) - f(a)}{2 - a}[/tex] as a approaches 2. Note that this new function is a function in a, i.e., it takes an input value a and returns the slope of secant line passing though the value of f at 2 and the value of f at a. So the limit of this new function as a approaches 2 is equal to the slope of the tangent line.

Here's how to work out the answer, so don't read this if you want to finish working it out on your own:






[tex]\lim_{a\rightarrow 2}\frac{f(2) - f(a)}{2 - a}[/tex]

[tex]= \lim_{a\rightarrow 2}\frac{4 - 2^{2} - 2a + a^{2}}{2 - a}[/tex]

[tex]= \lim_{a\rightarrow 2}\frac{-2a + a^{2}}{2 - a}[/tex]

[tex]= \lim_{a\rightarrow 2}\frac{(2 - a)(-a)}{2 - a}[/tex]

[tex]= \lim_{a\rightarrow 2}-a[/tex]

[tex]= -2[/tex]
 
  • #6
445
0
Wow, thank you all so much for all your help. With said help I was able to complete all the exercise problems in the textbook. Thanks again.
 

Related Threads for: Beginner(self study) calculus question.

  • Last Post
Replies
1
Views
639
  • Last Post
Replies
2
Views
1K
Replies
26
Views
4K
Replies
4
Views
3K
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
Top