Answer
Result: f'(-1)= -2
Work Step by Step
Step-1:
The derivative f'(a) is defined by the following equivalent limits.
$ f'(a) = \lim\limits_{h \to 0} f(a+h) - f(a)\div h $ ......(1)
And $ f'(a)= \lim\limits_{x \to a} f(x)-f(a)\div x-a $ ....(2)
Step-2:
Let f(x) = 3x^{2} +4x +2 , a= -1
Using (1) we have,
f'(-1) = \lim\limits_{h \to 0} f(-1+h) -f(-1)\div h
= \lim\limits_{h \to 0} (-1+h)^{2} +(-1+h) +2-[3(-1)^{2} +4(-1) +2 \div h
= \lim\limits_{h \to 0} 3-6h +3h^{2} -4+4h+1\div h
=\lim\limits_{h \to 0} 3h^{2} -2h\div h
\lim\limits_{h \to 0} 3h-2=-2
Now, using (2) we have,
f'(-1)=\lim\limits_{x \to -1} f(x)-f(-1)\div x-(-1)
= \lim\limits_{x \to -1} 3x^{2} +4x+2-1\div x+1
= \lim\limits_{x \to -1} 3x^{2} +4x+1\div x+1
= \lim\limits_{x \to -1} (3x+1)(x+1)\div(x+1)
= \lim\limits_{x\to -1} (3x+1)= -2
