- #1
danielle36
- 29
- 0
Hello,
I'm having trouble understanding an example from my test and I would appreciate your help clarifying how to get from one step to the next.
Problem: Find an equation of the tangent line to the hyberola y = 3/x at the point (3,1)
Eqation: [tex] m = \frac{f(a+h) - f(a)}{h} [/tex]
[tex] m = \frac{f(3 + h) - f(3)}{h} [/tex]
[tex]= \frac{\frac {3}{3 + h} - 1}{h} [/tex]
[tex]= \frac{\frac {3 - (3 + h)}{3 + h}}{h} [/tex]
(I'm assuming here they have just converted 1 to 3 + h / 3 + h, but I don't understand how they get to the next step...)
[tex]
= \frac{-h}{h(3+ h)}[/tex]
[tex]= \frac{-1}{3 + h} [/tex]
[tex]= \frac{-1}{3}[/tex][/tex]
I'm having trouble understanding an example from my test and I would appreciate your help clarifying how to get from one step to the next.
Problem: Find an equation of the tangent line to the hyberola y = 3/x at the point (3,1)
Eqation: [tex] m = \frac{f(a+h) - f(a)}{h} [/tex]
[tex] m = \frac{f(3 + h) - f(3)}{h} [/tex]
[tex]= \frac{\frac {3}{3 + h} - 1}{h} [/tex]
[tex]= \frac{\frac {3 - (3 + h)}{3 + h}}{h} [/tex]
(I'm assuming here they have just converted 1 to 3 + h / 3 + h, but I don't understand how they get to the next step...)
[tex]
= \frac{-h}{h(3+ h)}[/tex]
[tex]= \frac{-1}{3 + h} [/tex]
[tex]= \frac{-1}{3}[/tex][/tex]