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How do you use the difference quotient to find the derivative of fx 7 4X-2x2 where x3?
Wiki User
∙ 14y ago
f(x) = 7 + 4x - 2x2
Difference Quotient
You are going to plug in f(x) into:
f '(x) lim Δx -> 0 = f(x + Δx) - f(x) / Δx
[7 + 4(x + Δx) - 2(x + Δx)2] - (7 + 4x - 2x2) / Δx
= 7 + 4x + 4Δx - 2x2 - 4xΔx - 4(Δx)2 - 7 - 4x + 2x2 / Δx
= 4Δx - 4xΔx - 4(Δx)2 / Δx
= Δx (4 - 4x - 4Δx) / Δx
= (4 - 4x - 4Δx)
= 4 - 4x - 4(0)
= 4 - 4x
You can check using Power Rule:
f(x) = 7 + 4x - 2x2 f'(x) = 0 + 4(1)x1-1 - 2(2)x2-1
= 4 - 4x
When x = 3:
f'(3) = 4 - 4(3)
= 4 - 12
= -8
Wiki User
∙ 14y ago
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