1. If $\displaystyle f(x)=\frac{3x^2-5}{x+6}$ then f(0) + f'(0) is ....
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?

2.

3. Originally Posted by Monoxdifly
If $\displaystyle f(x)=\frac{3x^2-5}{x+6}$ then f(0) + f'(0) is ....
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?
Your calculation is correct, and the answer is not one of the listed options. Maybe you should check whether you read the question correctly.

4. Yes, the correct answer is $\displaystyle -\frac{25}{36}$.