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

#### Monoxdifly

##### Well-known member
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

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Helper
Yes, the correct answer is $$-\frac{25}{36}$$.

#### Monoxdifly

##### Well-known member
OK, thanks for the clarifications...