How do you use the definition of a derivative to find the derivative of #f(x)=x^3 + 2x^2 + 1#, at c=-2?

Question

Isaiah Allen · Accepted Answer

4

f'(-2) = #lim h->0 (f(-2+h) -f(-2))/h# = #lim h->0 ((-2+h)^3 +2(-2+h)^2 +1 - ( -8+8+1))/h # = #lim h->0 (-8+12h-6h^2 +h^3 +2(4-4h+h^2 )+1 -1)/h# =#lim h->0 (h^3 -4h^2 +4h)/h # = 4

Emily Ammerman · Answer

undefined To find the derivative of $ f(x) = x^3 + 2x^2 + 1 $ at $ c = -2 $, use the definition of a derivative:

$ f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} $

Substitute $ f(x) $ into the definition:

$ f'(c) = \lim_{h \to 0} \frac{((c + h)^3 + 2(c + h)^2 + 1) - (c^3 + 2c^2 + 1)}{h} $

Expand and simplify the expression:

$ f'(c) = \lim_{h \to 0} \frac{(c^3 + 3c^2h + 3ch^2 + h^3 + 2c^2 + 4ch + 2h^2 + 1) - (c^3 + 2c^2 + 1)}{h} $

$ f'(c) = \lim_{h \to 0} \frac{3c^2h + 3ch^2 + h^3 + 4ch + 2h^2}{h} $

Factor out $ h $ from the numerator:

$ f'(c) = \lim_{h \to 0} (3c^2 + 3ch + h^2 + 4c + 2h) $

Evaluate the limit as $ h $ approaches 0:

$ f'(c) = 3c^2 + 0 + 0 + 4c + 0 $

Now, substitute $ c = -2 $ into the expression:

$ f'(-2) = 3(-2)^2 + 4(-2) $

$ f'(-2) = 12 - 8 $

$ f'(-2) = 4 $

Therefore, the derivative of $ f(x) = x^3 + 2x^2 + 1 $ at $ c = -2 $ is $ f'(-2) = 4 $.