Caculus Help

[davidh93]

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

 

[Koot]

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.

 

[davidh93]

: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

 

[Koot]

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

 

[davidh93]

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