Power Rule Question?

Question

Can someone help me with this problem?

Using the Power Rule, show that the slope of the graph y = sqrt(x) at any point x is NOT halfway between the slope of the forward secant on [x, x + h] and the slope of the backward secant on [x−h, x].

Answer 1

The slope of the forward secant from x to x+h is:

(√(x+h) - √x)/h

And the slope of the reverse is:

(√(x-h) - √x)/(-h) = (√x - √(x-h))/h

The average of these two slopes is:

((√(x+h) - √(x-h))/h)/2 = (√(x+h) - √(x-h))/(2h)

Conversely, the derivative of √x at x is 1/(2√x) (it follows since d(√x)/dx = d(x^(1/2))/dx = 1/2x^(-1/2) = 1/(2√x)). So we have to show that 1/(2√x) ≠ (√(x+h) - √(x-h))/(2h) for any value of x and h (where obviously h≠0). Suppose the contrary, that 1/(2√x) = (√(x+h) - √(x-h))/(2h). Then we have:

1/(2√x) = (√(x+h) - √(x-h))/(2h)

Multiplying both sides by √(x+h) + √(x-h):

(√(x+h) + √(x-h))/(2√x) = ((x+h) - (x-h))/(2h)
(√(x+h) + √(x-h))/(2√x) = 2h/(2h)
(√(x+h) + √(x-h))/(2√x) = 1
√(x+h) + √(x-h) = 2√x

Squaring both sides:

(x+h) + 2√(x+h)√(x-h) + (x-h) = 4x
2x + 2√(x+h)√(x-h) = 4x
2√(x+h)√(x-h) = 2x
√(x+h)√(x-h) = x
(x+h)(x-h) = x²
x²-h² = x²
-h² = 0
h = 0

Since h is in the denominator of the expression (√(x+h) - √(x-h))/(2h), h cannot be 0, so our initial assumption that (√(x+h) - √(x-h))/(2h) = 1/(2√x) is false. Therefore, the slope of the tangent line is not the average of the slope of the secants. Q.E.D.