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
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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
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
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