Finding the slope of f(x)=x^3-2 at a general point

[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 of$$f(x)=x^3-2$$ 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