[ASK] Derivative of an Algebraic Fraction find f(0) + f'(0)

[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?

 

[Opalg]

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.

 

[HallsofIvy]

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

 

[Monoxdifly]

OK, thanks for the clarifications...