How Do You Calculate the Slope of a Curve at a Point?

[iamsmooth]

Homework Statement

Find the slope of the curve y=\frac{x}{3x+2} at the point x = -2.

Homework Equations

<br /> \lim_{h \rightarrow 0}\frac{f(x_{0} + h) - f(x_{0})}{h} = m<br />

The Attempt at a Solution

If x = -2, then y = 1/2. I'm not sure what to do from here.

This is the first step, but I don't get how you obtain this:
<br /> \lim_{h \rightarrow 0}\frac{\frac{-2+h}{3(-2+h)+2}-\frac{1}{2}}{h}<br />

I just need an explanation of what we're doing here and why?

Thanks in advance for the help!

 

[Office_Shredder]

This would be very difficult to get from the definition of a limit. You probably have something called the quotient rule that will help greatly

 

[iamsmooth]

The differentiation rules aren't introduced for another one and a half chapters in my textbook. This part is introducing a definition to the slope of a curve. It says:

The slope of a curve C at a point P is the slope of the tangent line to C at P if such a tangent line exists. In particular, the slope of the graph of y = f(x) at the point x₀ is
<br /> \lim_{h \rightarrow 0}\frac{f(x_{0} + h) - f(x_{0})}{h} = m

And after this definition, it gives the example that I posted. I have the answer, just don't understand any of it :(

 

[Pengwuino]

This can be done with some algebraic manipulation. Let your first term in the numerator be \frac{-2+h}{3h-4} and get the second term of 1/2 into that same form so that you can take 3h-4 under the denominator of the whole and go from there.

 

[HallsofIvy]

f(x)= \frac{x}{3x+ 2}
so f(x_0+h)= \frac{x_0+h}{3(x_0+h)+ 2}= \frac{x_0+h}{3x_0+3h+2}

f(x_0+h)- f(x_0)= \frac{x_0+h}{3(x_0+h)+ 2}= \frac{x_0+h}{3x_0+3h+2}- \frac{x}{3x+ 2}

The "common denominator" is (3x_0+ 2)(3x_0+ 3h+ 2). Multiplying numerator and denominator of the first fraction by 3x_0+ 2 and the numerator and denominator of the second fraction by 3x_0+ 3h+ 2,

\frac{(x_0+h)(3x_0+ 2)}{(3x_0+ 3h+ 2)(3x_0+ 2)}- \frac{(x_0)(3x_0+ 3h+ 2)}{(3x_0+ 3h+ 2)(3x_0+ 2)}

Multiply out the products in the numerators. You can leave the denominators as they are:
\frac{3x_0^2+ 2x_0+ 3hx_0+ 2h}{(3x_0+ 3h+ 2)(3x_0+ 2)}- \frac{3x_0^2+ 3hx_0+ 2x_0}{(3x_0+ 3h+ 2)(3x_0+ 2)}

Now you see that the "3x_0^2", "3hx_0", and "2x_0" terms cancel leaving
\frac{2h}{(3x_0+ 3h+ 2)(3x_0+ 2)}

That is f(x_0+ h)- f(x_0). To form the "difference quotient" divide by h:
\frac{f(x_0+ h)- f(x_0)}{h}= \frac{2h}{h(3x_0+ 3h+ 2)(3x_0+ 2)}= \frac{2}{(3x_0+ 3h+ 2)(3x_0+ 2)}.

Finally, take the limit as h goes to 0. Since setting h to 0 does not make the denominator 0, you can do that simply by setting h= 0.