Adv. Functions: Rate of Change in Rational Functions

[harujina]

Homework Statement

m_tan for f(x) = 5/√ 3x ... at x=1

Homework Equations

m_sec = 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)

m_sec = 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...

 

[Mark44]

Homework Statement

m_tan for f(x) = 5/√ 3x ... at x=1

Please clarify what you're saying here. Which if these is your function?
$$1. f(x) = \frac{5}{\sqrt{3}}x$$
$$2. f(x) = \frac{5}{\sqrt{3x}}$$
$$3. f(x) = \frac{5}{\sqrt{3}x}$$

Homework Equations

m_sec = 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)

m_sec = 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...

 

[harujina]

Please clarify what you're saying here. Which if these is your function?
$$1. f(x) = \frac{5}{\sqrt{3}}x$$
$$2. f(x) = \frac{5}{\sqrt{3x}}$$
$$3. f(x) = \frac{5}{\sqrt{3}x}$$

Oh, the second one.
$$2. f(x) = \frac{5}{\sqrt{3x}}$$
Sorry about that!

 

[Mark44]

Homework Statement

m_tan for f(x) = 5/√ 3x ... at x=1

Homework Equations

m_sec = 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)

You need more parentheses.
The second point should be written as (1 + h, 5/√(3 + h)), otherwise the 2nd coordinate looks like this:
$$ \frac{5}{\sqrt{3}} + h$$

m_sec = f(1+h) - f(1) / h

The right side should be written as (f(1 + h) - f(1))/h. Most would interpret what you wrote as
$$ f(1 + h) - \frac{f(1)}{h}$$

= (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)

You have the right idea, but with all the missing parentheses, it's just too hard to read.

= 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...

 

[Mark44]

When it's completely simplified, you should get this:
$$ \frac{-5\sqrt{3}}{6}$$

BTW, LaTeX makes it much easier to read, and isn't that hard to write.
To create a fraction, do this:

\frac{a + b}{c}

I don't show the pairs of $ symbols at front and back.
This renders like so:
$$\frac{a + b}{c}$$

To create a square root, do this:

\sqrt{3 + h}

Again, you need $ pairs at front and back.
This is how it looks in the browser:
$$\sqrt{3 + h}$$