Use the alternate form of the derivative to find the derivative at x=c, if it exists.

Question

F(x)=x³+2x²+1, c=-2

Answer 1

f(x) = x³ + 2x² + 1

f'(c) = lim_x→c [ (f(x) - f(c)) / (x - c) ]

So, f'(-2) = lim_x→-2 [ (x³ + 2x² + 1 - 1 ) / (x + 2) ] = lim_x→-2 [ (x³ + 2x²) / (x + 2) ]

= lim_x→-2 [ x²(x + 2) / (x + 2) ] = lim_x→-2 [ x² ] = (-2)² = 4

Answer 2

The alternate form of the derivative is f'(c)=[lim as x→c] (f(x)-f(c))/(x-c). For your problem, f(x)=x³+2x²+1, and c= -2. f(c) would be equivalent to f(-2), which is (-2)³+2(-2)²+1. Evaluating this gives us -8+8+1=1. Let's plug these in.

f'(-2)=[lim as x→-2] ((x³+2x²+1)-(1))/(x-(-2))

Let's simplify.

= [lim as x→-2] (x³+2x²)/(x+2)

Apply the limit by plugging in -2 for every x in the function.

= ((-2)³+2(-2)²)/(-2+2)

This gives us 0/0, meaning we have to apply L'Hopital's Rule. This rule states that we must take the derivative of both the numerator and denominator of the function.

Using power rule, this would give us [lim as x→-2] (3x²+4x)/(1).

Plugging in -2 to evaluate the limit now gives us 3(-2)²+4(-2) which equals 12-8 = 4.

The derivative at x=c must therefore be 4. Hope this helps!