How do you find f'(-2) using the limit definition given f(x)=x^3+ 3xf(x)=x3+3x?

1 Answer
Steve M
Nov 29, 2016

f'(-2)=15

Explanation:

By definition of the derivative f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h
So with f(x) = x^3 + 3x we have;

f'(-2)=lim_(h rarr 0) ( {(-2+h)^3+3(-2+h) } - {(-2)^3+3(-2) } ) / h
:. f'(-2)=lim_(h rarr 0) ( {(-2)^3+3(-2)^2h+3(-2)h^2+h^3-6+3h } - {-8-6 } ) / h
:. f'(-2)=lim_(h rarr 0) (-8+12h-6h^2+h^3-6+3h +14 ) / h
:. f'(-2)=lim_(h rarr 0) (15h-6h^2+h^3 ) / h
:. f'(-2)=lim_(h rarr 0) 15-6h+h^2
:. f'(-2)=15

NB;
Compare with f'(x)=3x^2+3 => f'(-2)=12+3=15

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