Sunday, February 3, 2013

Find f '(c) by forming the difference quotient [f(c+h)-f(c)]/h and taking the limit as h=>0. f(x)=2x^3 +1 ; c=1

To calculate the value of the first derivative in a given
point, c = 1, we'll have to apply the limit of the
ratio:


limit [f(x) - f(1)]/(x-1), when x tends to
1.


We'll substitute f(x) and we'll calculate the value of
f(1):


f(1) = 2*1^3 + 1


f(1) =
2+1


f(1) = 3


limit [f(x) -
f(1)]/(x-1) = lim (2*x^3 + 1 - 3)/(x - 1)


We'll combine
like terms:


lim (2*x^3 + 1 - 3)/(x - 1) = lim (2*x^3 -
2)/(x - 1)


We'll factorize the numerator by
2:


lim (2*x^3 - 2)/(x - 1) = lim
2(x^3-1)/(x-1)


We'll write the difference of cubes as a
product:


x^3 - 1 = (x-1)(x^2 + x +
1)


lim 2(x^3-1)/(x-1) = 2 lim (x-1)(x^2 + x +
1)/(x-1)


We'll simplify the ratio and we'll
get:


2 lim (x-1)(x^2 + x + 1)/(x-1) = 2 lim (x^2 + x +
1)


We'll substitute x by 1 and we'll
get:


2 lim (x^2 + x + 1) = 2(1^2 + 1 +
1)


2 lim (x^2 + x + 1) = 2*3


2
lim (x^2 + x + 1) = 6


But f'(c) = f'(1) = limit [f(x) -
f(1)]/(x-1)


f'(1) =
6

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