Find f'(c) for c = 2 and c = -1 forming the difference quotient f(x)= 2x + 3

To calculate the value of the first derivative in a given
point, c = 2, we'll have to apply the limit of the difference
quotient:

limit [f(x) - f(2)]/(x-2), when x tends to c =
2.

We'll substitute f(x) and we'll calculate the value of
f(2):

f(2) = 2*2 + 3

f(2) =
4+3

f(2) = 7

limit [f(x) -
f(2)]/(x-2) = lim (2*x + 3 - 7)/(x - 2)

We'll combine like
terms:

lim (2x - 4)/(x -
2)

We'll factorize the numerator by
2:

lim (2x - 4)/(x - 2) = lim
2(x-2)/(x-2)

lim 2(x-2)/(x-2) =
2

But f'(c) = f'(2) = limit [f(x) -
f(2)]/(x-2)

f'(2) =
2

Now, we'll calculate the value of the
first derivative in a given point, c = -1,

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

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

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

f(-1) = 1

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

We'll combine like
terms:

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

We'll factorize the numerator by
2:

lim (2x+2)/(x+1)  = lim
2(x+1)/(x+1)

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

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

f'(-1) =
2