Power Rule Question?

Question

Can someone explain to me how to do this problem?

Using the Power Rule, show that the slope of the graph y = x2 at any point x is exactly halfway between the slope of the forward secant on [x, x + h] and the slope of the backward secant on [x−h, x].

Why is this never the case for y = x^3 and y = sqrt(x)

Answer 1

I'll show you how to do it with y=x^3, and then you can do the same thing with y=x^2 and y=sqrt(x) and see what results you get.

Use the power rule on the curve y=x^3
The derivative is y'=3x^2

Now find the slope of the forward secant, which is a line which passes through two points of the curve: the first point is (x,y) and we know y=x^3 so the point is really (x,x^3) and the second point is (x+h,___). We have to figure out what goes in that blank. To do this, plug "x+h" into the y=x^3 formula where the x used to be: y=(x+h)^3. This equals x^3+3x^2h+3xh^2+h^3, so the second point is (x+h,x^3+3x^2h+3xh^2+h^3)

So, what's the slope of the line passing through those two points? It's the difference in the y-coordinates divided by the difference in the x-coordinates.
slope = ( x^3+3x^2h+3xh^2+h^3 - x^3 ) / ( x+h - x )
= (3x^2h+3xh^2+h^3) / h
= 3x^2+3xh+h^2

Now do the same thing with the backward secant, which is a line which passes through two points of the curve: the first point is (x,x^3) and the second point is (x-h,___). What goes in that blank? Plug "x-h" into the y=x^3 formula and get x^3-3x^2h+3xh^2-h^3, so the second point is (x-h,x^3-3x^2h+3xh^2-h^3)

Hence, the slope is ( x^3-3x^2h+3xh^2-h^3 - x^3 ) / ( x-h - x )
= (-3x^2h+3xh^2-h^3) / -h
= 3x^2-3xh+h^2

Now find the number "halfway between" the two slopes (in other words, average them together) and see if the result is exactly equal to y'

(3x^2+3xh+h^2 + 3x^2-3xh+h^2) / 2
= (6x^2+2h^2) / 2
= 3x^2+h^2

Remember y'=3x^2.
Are these two answers identical? No.

Now repeat this whole long process for y=x^2.

And then do it again for y=sqrt(x)
Here's a hit: sqrt(x)=x^(1/2)

Answer 2

Show that the average of the slopes of the forward secant and the backward secant is 2x.

The reason this does not hold for x^3 or sqrt(x) is because the derivatives of those functions are not linear.