- Thread starter
- #1

#### Monoxdifly

##### Well-known member

- Aug 6, 2015

- 269

A. 2

B. 1

C. 0

D. -1

E. -2

What I did:

If \(\displaystyle f(x)=\frac{u}{v}\) then:

u =\(\displaystyle 3x^2-5\) → u' = 6x

v = x + 6 → v' = 1

f'(x) =\(\displaystyle \frac{u'v-uv'}{v^2}\)=\(\displaystyle \frac{6x(x+6)-(3x^2-5)(1)}{(x+6)^2}\)

f(0) + f'(0) = \(\displaystyle \frac{3(0^2)-5}{0+6}\) + \(\displaystyle \frac{6(0)(0+6)-(3(0^2)-5)(1)}{(0+6)^2}\) = \(\displaystyle \frac{3(0)-5}{6}\) + \(\displaystyle \frac{0(0+6)-(3(0)-5)}{6^2}\)= \(\displaystyle \frac{0-5}{6}\) + \(\displaystyle \frac{0-(0-5)}{36}\) = \(\displaystyle \frac{-5}{6}\) + \(\displaystyle \frac{0-(-5)}{36}\) = \(\displaystyle \frac{-30}{36}\) + \(\displaystyle \frac{0+5}{36}\) = \(\displaystyle \frac{-25}{36}\)

The answer isn't in any of the options. I did nothing wrong, right?