- #1
harujina
- 77
- 1
Homework Statement
mtan for f(x) = 5/√ 3x ... at x=1
Homework Equations
msec = y2-y1 / x2-x1
The Attempt at a Solution
The two points I got from the equation: (1, 5/√ 3) and (1+h, 5/√ 3+h)
msec = f(1+h) - f(1) / h
= (5/√ 3+h - 5/√ 3) / h ... multiply top and bottom by denominators (√ 3+h) and (√ 3)
= 5√ 3 - 5√ 3+h / h(√ 3+h)(√ 3) ... multiply top and bottom by the conjugate (5√ 3 + 5√ 3+h)
= 25√ 9 + 25√ 9+h - 25√ 9+h - 25√ 9+h2 / h(√ 3+h)(√ 3)(5√ 3+5√ 3+h)
= 25√ 9 - 25√ 9+h2 / h(√ 9+h)(5√ 3+5√ 3+h)
= 75 - 25√ 9+h2 / h(5√ 27+3h + 5√ 27+3h2)
No idea if this is right and where I'm going with this.
I don't know the answer to this question either, so can't check my work...