If f(x) = x^2 + 2x, evaluate f '(a) using equation (1) of Theorem 3.5

i don't know if this is what you want . . .

x^2 + 2x = (a^2 + 2a) + (x^2 + 2x) - (a^2 + 2a)
. . . . . . = (a^2 + 2a) + [ (x^2 - a^2) + (2x - 2a) ]
. . . . . . = (a^2 + 2a) + (x -a) (x+a+2)
. .. . . . .= f(a) + (x -a) * F(x)

Then limit F(x) as x ---> a = limit (x + a+2) as x ---> a = 2a + 2

hence, f '(a) = 2a + 2 . . .answer

You can't just tell us to use equation (1) of Theorem 3.5
This only has relevance if we are using the same textbook as you are

f(x) = x² + 2x
f'(x) = 2x + 2
f'(a) = 2a + 2

Or perhaps you need to use limit definition of derivative
(this makes sense to everyone, and does not require us to have same textbook)

f'(a) = lim[h→0] (f(a+h) − f(a)) / h
. . . . = lim[h→0] ((a+h)² + 2(a+h) − (a² + 2a)) / h
. . . . = lim[h→0] (a² + 2ah + h² + 2a + 2h − a² − 2a) / h
. . . . = lim[h→0] (2ah + h² + 2h) / h
. . . . = lim[h→0] h (2a + h + 2) / h
. . . . = lim[h→0] (2a + h + 2)
. . . . = 2a + 0 + 2
. . . . = 2a + 2

Another method using limits is:

f'(a) = lim[x→a] (f(x) − f(a)) / (x−a)
. . . . = lim[x→a] ((x² + 2x) − (a² + 2a)) / (x−a)
. . . . = lim[x→a] ((x² − a²) + (2x − 2a)) / (x−a)
. . . . = lim[x→a] ((x − a)(x + a) + 2 (x − a)) / (x−a)
. . . . = lim[x→a] (x − a) (x + a + 2) / (x−a)
. . . . = lim[x→a] (x + a + 2)
. . . . = a + a + 2
. . . . = 2a + 2

P.S. Please use general terminology when asking questions, instead of terminology specific to a particular textbook. I still don't know if I've answered correctly.

666

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