Finding equation to tangent line

[danielle36]

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: m = \frac{f(a+h) - f(a)}{h}


m = \frac{f(3 + h) - f(3)}{h}

= \frac{\frac {3}{3 + h} - 1}{h}

= \frac{\frac {3 - (3 + h)}{3 + h}}{h}

(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...)

<br /> = \frac{-h}{h(3+ h)}

= \frac{-1}{3 + h}

= \frac{-1}{3}[/tex]

 

[HallsofIvy]

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: m = \frac{f(a+h) - f(a)}{h}


m = \frac{f(3 + h) - f(3)}{h}

= \frac{\frac {3}{3 + h} - 1}{h}

= \frac{\frac {3 - (3 + h)}{3 + h}}{h}

(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...)

Yes,
\frac{\frac{3}{3+h}- 1}{h}=\frac{\frac{3}{3+h}- \frac{3+h}{3+h}}{h}= \frac{\frac{-h}{3+h}}{h}
And dividing by h is the same as multiplying by 1/h
\frac{\frac{-h}{3+h}}{h}= \frac{-h}{3+h}\frac{1}{h}= \frac{-h}{h(3+h)}

<br /> = \frac{-h}{h(3+ h)}

= \frac{-1}{3 + h}

By the way, this last equality is true as long as h is not equal to 0 since in that case we have 0/0. Fortunately, the limit, as h goes to 0 does not depend on the value at 0.

= \frac{-1}{3}

 

[danielle36]

Ohh right! Thanks! That's what I figured was going on but I was't keeping the brackets in the denominator, I don't know why I couldn't see that now!

 

[danielle36]

Ok, so I was going over this example again, and I've realized I'm not sure what is going on in that last step (where the last h is gone, seemingly at random)... Has 0 been substituted for h?

 

[HallsofIvy]

The limit as h goes to 0. That is not the same as "0 substituted for h", especially here because, as I said before "this last equality is true as long as h is not equal to 0 since in that case we have 0/0."
But I also said "Fortunately, the limit, as h goes to 0, does not depend on the value at 0". Since, as long as h is not o, this is the same as -1/(3+h) and that is continuous at h= 3, the limit can be calculated by setting h= 0 in that: -1/(3+ 0)= -1/3.

 

[danielle36]

Hmm I'm not sure if I know what you mean, because I think I have fallen into a similar situation with a problem I was working on...

The solution in this case is 2..
Find the slop of the tangent line to the parabola y = 4x - x^{2} at the point (1,3)

m = \frac{f(x+h) - f(x)}{h}
= \frac{(4[1+h] - [1 + h]^{2}) - 3)}{h}
= \frac{4 + 4h - 1 - 2h -h^{2} - 3}{h}
= 2 - h

Then I'm not sure what to do with the h... so maybe this is the same situation? When can h not be set at 0?

 

[HallsofIvy]

By the time you are looking at the definition of "derivative" and trying to use it, you should already be familiar with "limits" and "continuity". Do you remember the definition of "\lim_{h\rightarrow 0} f(h)&quot; and the condition under which it is equal to f(0)?

 

[danielle36]

Yeah I'm familiar with the little limit symbol with the right arrow under it, obviously not familiar enough though... I do know where to find more information on it in my text though, thanks for showing me where to look!

 

[HallsofIvy]

Hmm I'm not sure if I know what you mean, because I think I have fallen into a similar situation with a problem I was working on...

The solution in this case is 2..
Find the slop of the tangent line to the parabola y = 4x - x^{2} at the point (1,3)

m = \frac{f(x+h) - f(x)}{h}
= \frac{(4[1+h] - [1 + h]^{2}) - 3)}{h}
= \frac{4 + 4h - 1 - 2h -h^{2} - 3}{h}
= 2 - h

Then I'm not sure what to do with the h... so maybe this is the same situation? When can h not be set at 0?

I wasn't asking if you were familiar with "the little symbol", but with the basic properties of limits and continuity!

= \frac{4 + 4h - 1 - 2h -h^{2} - 3}{h}= \frac{2h- h^2}{h}
= \frac{h(2- h)}{h}
and, for h not equal to 0, the hs cancel out.
Now take the limit as h goes to 0. If h is "very, very, very" close to 0, then obviously 2- h is "very, very, very" close to 2- 0 and the limit is 2.

Recall that a function is said to be "continuous at a" precisely if \lim_{x\rightarrow a} f(x)= f(a). You should also be aware of theorems that tell you &quot;all polynomials are continuous for all a&quot;, &quot;all rational functions are continuous whenever the denominator is not 0&quot;. Taking the limit as h goes to 0 is NOT just &quot;setting h= 0&quot; but in those situations, or situations you can reduce to those, they <b>are</b> the same.