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Caculus Help
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  #1  
Old 10-07-2010, 11:06 PM
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Default Caculus Help

Hey guys,

I need some help on my calc homework. I have no idea what's going on

Find the derivative for each function in exercises 1-12. Use the definition:

For exercises 13- 17, leave the equations you write in slope-intercept form. In exercises 18 - 20 simplify all fractions to lowest terms but you should leave fractions in improper form ( that is the numerator can be larger than the denominator like 9/5 but the numbers cannot share a GCF. So 8/6 should become 4/3) .
  1. f (x) = - 6x
  2. f (x) = 4x + 2
  3. f (x) = 3x - 10
  4. f (x ) = 2x
  5. f (x) = x2 - 6
  6. f (x) = 4x2 + 1
  7. f (x ) = - 5x2
  8. f (x ) = x2 + 4x
  9. f (x ) = 2x2 - 3x
  10. f (x ) = x3 - 1
  11. f (x ) = x2 + x + 1
  12. f (x) = 2x - x3
  13. Find the equation of the line tangent to the graph of f(x) = 2x2 + 1 at the point where x = 1
  14. Find the equation of the line tangent to the graph of f(x) = 2x2 + 1 at the point where x = 2
  15. Find the equation of the line tangent to the graph of f(x) = x3 - 2 at the point where x = 0
  16. Find the equation of the line tangent to the graph of f(x) = 2x + 1 at the point where x = - 3.
  17. Find the equation of the line tangent to the graph of f(x) = 8 at the point where x = 3.
  18. For the function f ( x ) = x2 + 1, find the slope of the secant line between the points (0 , f(0)) and (3 , f (3)).
  19. For the function f ( x ) = 2x2 - 3x, find the slope of the secant line between the points (-1 , f (-1 )) and (0 , f(0)).
  20. For the function f (x ) = 1 - x3 , find the slope of the secant line between the points (3, f(3)) and (5 , f(5)).

any help would be great because i have no clue how to do derivatives

edit: so i figured out 1 -12, but the rest is still very annoying. and got most of the tangent lines, but still need help on secant

Last edited by davidh93 : 10-07-2010 at 11:37 PM.
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  #2  
Old 10-07-2010, 11:38 PM
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Default Re: Caculus Help

If you have no idea how to do derivatives then I assume you don't know about the power rule which is the fastest way of doing them.

Since you probably don't you'll have to use the limit

lim ( f(x+h) - f(x) ) / h
h->0

Just plug in the equations for f(x), solve the limit and the answer will be your derivative.

Last edited by Koot : 10-07-2010 at 11:38 PM.
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Old 10-07-2010, 11:52 PM
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Default Re: Caculus Help

:O thanks i know how to do derivatives and semi how to find the tangent line, but on the secants i just bsed... I am already late on this assignment so it doesnt really matter. Thanks anyways
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Old 10-07-2010, 11:55 PM
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Default Re: Caculus Help

Well, just to learn the secant lines. It's basically a line going through the curve and you have the two points there.

The f(x) can be used to find the number, then the simple slope formula can be used.

y2 - y1 / x2 - x1
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  #5  
Old 10-08-2010, 07:53 AM
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Default Re: Caculus Help

:O thanks for your help, mods u can close this
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