Slopes and Tangent Lines Question

[Silvanoshei]

Gotta find g(x)=x/x-2 at point (3,3). Pretty easy to find f(3) which is 3/3-2 = 3.

f(h+3) is giving me problems though. When @ x=h+3, it's (h+3)/(h+3)-2. Which wouldn't the (h+3) factor out? And give us -1/2?

I'm doing something wrong b/c f(h+x)-(x)/h should factor nicely. -1/2-3/h is not nice. Division is giving me a hick up here.

 

[DrPhil]

Always use "extra" parentheses when typing in line, to show the order of operations.

Gotta find g(x)=x/(x-2) at point (3,3). Pretty easy to find f(3) which is 3/(3-2) = 3.

f(h+3) is giving me problems though. When @ x=h+3, it's (h+3)/((h+3)-2). Which wouldn't the (h+3) factor out? And give us -1/2?

NO, you can't factor out (h+3) because (h+3) is not a factor of 2. The fraction is (h+3)/(h+1)

I'm doing something wrong b/c (f(h+x)-f(x))/h should factor nicely. -1/2-3/h is not nice. Division is giving me a hick up here.

f(h+3) - f(3) = (h+3)/(h+1) - 3

Combine the two terms over the common denominator (h+1). Then divide by h.

 

[Silvanoshei]

I see. So it would be (h+3)(h-1) / (h+1)(h-1), ending up with h²-2h-3 -3 / h ...?

 

[JeffM]

I see. So it would be (h+3)(h-1) / (h+1)(h-1), ending up with h²-2h-3 -3 / h ...?

NO

\(\displaystyle g(x) = \dfrac{x}{x - 2}\ if\ x \ne 2 \implies g(3) = \dfrac{3}{3 - 2} = 3.\)

\(\displaystyle g(3 + h) = \dfrac{3 + h}{3 + h - 2} = \dfrac{3 + h}{h + 1}.\)


\(\displaystyle g(3 + h) - g(3) = \dfrac{3 + h}{h + 1} - 3 = \dfrac{3 + h - 3(h + 1)}{h + 1} = \dfrac{-2h}{h + 1}.\)

Now continue.

 

[Silvanoshei]

So you take the -3 times the denominator and stick it up in the numerator? What's this called?

With that, (-2h/h+1) / h is simply -2/1 or m=-2.

 

[srmichael]

So you take the -3 times the denominator and stick it up in the numerator? What's this called?

With that, (-2h/h+1) / h is simply -2/1 or m=-2.

JeffM is simply showing how you get a common denominator in order to subtract the two terms. The common denominator is h + 1 thus you multiply the 3 by (h+1)/(h+1) which is equivalent to 1 thus you are not changing the problem, yet you are getting a common denominator that allows you to subtract the fractions.

Perhaps this may make it a little clearer to you:

\(\displaystyle \frac{3+h}{h+1}-3=\frac{3+h}{h+1}-\frac{(3)(h+1)}{h+1}=\frac{(3+h)-3(h+1)}{h+1}=\frac{3+h-3h-3}{h+1}=\frac{-2h}{h+1}\)

 

[mmm4444bot]

Division is giving me a hick up

It seems like you need to review working with algebraic fractions.

Cheers :cool:

 

[JeffM]

So you take the -3 times the denominator and stick it up in the numerator? What's this called?

With that, (-2h/h+1) / h is simply -2/1 or m=-2.

Pause a moment.

First, you need to remember basic algebra. Calculus will kill you if you forget your algebra.

\(\displaystyle \dfrac{3 + h}{h + 1} - 3 = \dfrac{3 + h}{h + 1} - (3 * 1) = \dfrac{3 + h}{h + 1} - 3\left(\dfrac{h + 1}{h + 1}\right) = \dfrac{3 + h}{h + 1} + \dfrac{-3(h + 1)}{h + 1} = \dfrac{3 + h - 3h - 3}{h+1} = \dfrac{-2h}{h + 1}.\)

You are absolutely correct that:

\(\displaystyle \dfrac{\dfrac{-2h}{h + 1}}{h} = \dfrac{-2}{h + 1}.\)

How in the world do you conclude that \(\displaystyle \dfrac{-2}{h + 1} = - 2?\)

\(\displaystyle h = 1 \implies \dfrac{-2}{h + 1} = - 1.\)

What you mean is that

\(\displaystyle \displaystyle \lim_{h \rightarrow 0}\dfrac{-2}{h + 1} = \dfrac{\displaystyle \lim_{x \rightarrow 0}-2}{\displaystyle \lim_{h \rightarrow 0}(h + 1)} = \dfrac{-2}{1} = - 2.\)