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

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f(x) = 2x+3

Using thr difference quotient we know that:

f'(c) = lim [f(x) -f(c) ]/(x-c) x --> c

==> f'(2) = lim [(f(x) - f(2)]/(x-2)

= lim (2x+3 - 7 )/(x-2)

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

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

= lim 2 x--> 2

**==> f'(x) = 2**

**==> f'(2) = 2**

**==> f'(-1) = 2 **

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**

f(x) = 2x+3.

To find f'(c) for c = 2 and c = -1.

Solution:

f(x) = 2x+3.

Differentiating , we get:

f'(x) = (2x+3)' = (2x)' +(3)'

f'(x) - 2+0, as (kx^n) = k*nx^(n-1).

f(x) = 2. for all x.

f'(c) = 2 when c = 2

Or f '(2) = 2.

Also f '(c) = 2, when c = 1.

Or f '(1) = 2

So f(2) = f(1) = 2