SOLUTION: Given g(x) = 3x^2 - 2x + 1, find [g(a + h) - g(a)]/h. a. 6a + 3h + 1 b. 6a + 3h - 1 c. 6a + 3h + 2 d. 6a + 3h -2 e. none of these

Given g(x) = 3x^2 - 2x + 1, find [g(a + h) - g(a)]/h. 
a. 6a + 3h + 1
b. 6a + 3h - 1
c. 6a + 3h + 2
d. 6a + 3h -2
e. none of these

First find g(a + h) by plugging in a + h for x in

    g(x) = 3x² - 2x + 1
g(a + h) = 3(a + h)² - 2(a + h) + 1

g(a + h) = 3(a + h)(a + h) - 2a - 2h + 1

g(a + h) = 3(a² + 2ah + h²) - 2a - 2h + 1

g(a + h) = 3a² + 6ah + 3h² - 2a - 2h + 1 

Next find g(a) by plugging in a for x in

    g(x) = 3x² - 2x + 1

    g(a) = 3a² - 2a + 1

Nwxt plug (3a² + 6ah + 2h² - 2a - 2h + 1) for 
g(a + h) and (3a² - 2a + 1) for g(a) in

[g(a + h) - g(a)]/h 

[g(a + h) - g(a)]/h = [(3a² + 6ah + 2h² - 2a - 2h + 1) - (3a² - 2a + 1)]/h

[g(a + h) - g(a)]/h = [3a² + 6ah + 3h² - 2a - 2h + 1 - 3a² + 2a - 1]/h

Cancel whatever will add to 0:

[g(a + h) - g(a)]/h = [6ah + 3h² - 2h]/h

[g(a + h) - g(a)]/h = 6ah/h + 3h²/h - 2h/h

Cancel h's in fractions:

[g(a + h) - g(a)]/h = 6a + 3h - 2

So the answer is (d)

Edwin