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

AI Thread Summary
The discussion revolves around calculating f(0) + f'(0) for the function f(x) = (3x^2 - 5) / (x + 6). The calculations provided confirm that f(0) + f'(0) equals -25/36, which is not among the provided answer choices. Participants agree that the calculations are correct and suggest verifying the question's details. The thread highlights a potential oversight in the answer options given. The conclusion emphasizes that the calculated result does not match any listed answers.
Monoxdifly
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If $$f(x)=\frac{3x^2-5}{x+6}$$ then f(0) + f'(0) is ...
A. 2
B. 1
C. 0
D. -1
E. -2

What I did:
If $$f(x)=\frac{u}{v}$$ then:
u =$$3x^2-5$$ → u' = 6x
v = x + 6 → v' = 1
f'(x) =$$\frac{u'v-uv'}{v^2}$$=$$\frac{6x(x+6)-(3x^2-5)(1)}{(x+6)^2}$$
f(0) + f'(0) = $$\frac{3(0^2)-5}{0+6}$$ + $$\frac{6(0)(0+6)-(3(0^2)-5)(1)}{(0+6)^2}$$ = $$\frac{3(0)-5}{6}$$ + $$\frac{0(0+6)-(3(0)-5)}{6^2}$$= $$\frac{0-5}{6}$$ + $$\frac{0-(0-5)}{36}$$ = $$\frac{-5}{6}$$ + $$\frac{0-(-5)}{36}$$ = $$\frac{-30}{36}$$ + $$\frac{0+5}{36}$$ = $$\frac{-25}{36}$$
The answer isn't in any of the options. I did nothing wrong, right?
 
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Monoxdifly said:
If $$f(x)=\frac{3x^2-5}{x+6}$$ then f(0) + f'(0) is ...
A. 2
B. 1
C. 0
D. -1
E. -2

What I did:
If $$f(x)=\frac{u}{v}$$ then:
u =$$3x^2-5$$ → u' = 6x
v = x + 6 → v' = 1
f'(x) =$$\frac{u'v-uv'}{v^2}$$=$$\frac{6x(x+6)-(3x^2-5)(1)}{(x+6)^2}$$
f(0) + f'(0) = $$\frac{3(0^2)-5}{0+6}$$ + $$\frac{6(0)(0+6)-(3(0^2)-5)(1)}{(0+6)^2}$$ = $$\frac{3(0)-5}{6}$$ + $$\frac{0(0+6)-(3(0)-5)}{6^2}$$= $$\frac{0-5}{6}$$ + $$\frac{0-(0-5)}{36}$$ = $$\frac{-5}{6}$$ + $$\frac{0-(-5)}{36}$$ = $$\frac{-30}{36}$$ + $$\frac{0+5}{36}$$ = $$\frac{-25}{36}$$
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.
 
Yes, the correct answer is -\frac{25}{36}.
 
OK, thanks for the clarifications...
 
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