Use the limit definition of derivative to calculate the derivative of f(x)=x^2-4x+2?

Question

show your work

Answer 1

The limit definition of derivative is:

lim [f(x + Δx) - f(x)]/Δx = f'(x)
Δx→0

Since in your case, f(x) = x² - 4x + 2, just plug that in:

lim [(x + Δx)² - 4(x + Δx) + 2 - (x² - 4x + 2)]/Δx
Δx→0

Expand terms using FOIL:

lim [x² + 2xΔx + Δx² - 4x - 4Δx + 2 - x² + 4x - 2]/Δx
Δx→0

Simplify by removing the terms that cancel:

lim [2xΔx + Δx² - 4Δx]/Δx
Δx→0

Factor Δx from the numerator:

lim Δx[2x + Δx - 4)]/Δx
Δx→0

Cancel Δx in numerator and denominator, and then plug in Δx = 0:

lim [2x + Δx - 4)] = 2x - 4 = f'(x)
Δx→0

Answer 2

[f(x+h)-f(x)]/h = {[(x+h)^2 - 4(x+h) + 2] -(x^2 - 4x + 2)}/h =
= (x^2 + 2hx +h^2 - 4x - 4h + 2 - x^2 + 4x - 2)/h =
= (2hx + h^2 - 4h)/h = 2x +h - 4

Now, as h approaches 0 [f(x+h)-f(x)]/h = 2x+h-4 approaches 2x-4, which is defined to be f'(x)

Answer 3

f'(x) = lim(a->0) [f(x+a) - f(x)]/a
= lim(a->0) [(x+a)^2 - x^2 -4a]/a
= lim(a->0) [a^2 + 2ax -4a]/a
= lim(a->0) [a + 2x -4]
= 2x -4

Answer 2x -4