Help with Derivative Homework: Find Slope of f(x)=x^3-2

[priscilla98]

Homework Statement

Find the slope of the tangent to the graph of f (x) = x^3 - 2 at a general point Xo.

Homework Equations

lim = f (X1) - f (Xo)
X1 -> Xo -------------
X1 - Xo

The Attempt at a Solution

I can use the coordinates (2, 6)

lim = f (X1) - f (Xo)
X1 -> Xo -------------
X1 - Xo

= (x^3 -2) - (6)
--------------
x - 2

Am I going right with this? I would appreciate the help, thanks. And happy holidays.

 

[micromass]

Yes, you're on the right track. So the numerator is x^3-8. Can you factor this?

 

[Kevin_Axion]

Here I'll go through the process with you:

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

now your x here is x_0, so plugging in x_0 for x you get:

\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}

since your f(x)=x^3-2 you can evaluate f(x_0+h) and f(x_0) by plugging them into the x value in the original function namely, f(x)= x^3-2.

plugging in f(x_0+h) into the x value off(x)=x^3-2 we get f(x_0+h) = h^3+3h^2x_0+3h{x^2}_0+{x_0}^3-2.

Plugging in f(x_0) we get {x^3}_0-2 so our limit now looks like this:

\lim_{h \to 0} \frac{(h^3+3h^2x_0+3h{x^2}_0+{x_0}^3-2)-({x^3}_0-2)}{h} you will notice that the 2 and the {x^3}_0 cancel out and your left with
\lim_{h \to 0} \frac{(h^3+3h^2x_0+3h{x^2}_0)}{h} factoring out the h and dividing you end up with:

\lim_{h \to 0} {(h^2+3hx_0+3{x^2}_0)} and evaluating the limit you get that derivative at any point x_0 is

= 3{x_0}^2

I'm sorry if I misread the question, this is how I interpreted it.

 

[priscilla98]

No it's okay, thanks a lot for this. I'm usually confused with derivatives sometimes.

But thanks and happy holidays