
MHB Craftsman
#1
March 6th, 2020,
03:10
If $ \displaystyle f(x)=\frac{3x^25}{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^25$ → u' = 6x
v = x + 6 → v' = 1
f'(x) =$ \displaystyle \frac{u'vuv'}{v^2}$=$ \displaystyle \frac{6x(x+6)(3x^25)(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{05}{6}$ + $ \displaystyle \frac{0(05)}{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?

March 6th, 2020 03:10
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MHB Oldtimer
#2
March 6th, 2020,
08:37
Originally Posted by
Monoxdifly
If $ \displaystyle f(x)=\frac{3x^25}{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^25$ → u' = 6x
v = x + 6 → v' = 1
f'(x) =$ \displaystyle \frac{u'vuv'}{v^2}$=$ \displaystyle \frac{6x(x+6)(3x^25)(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{05}{6}$ + $ \displaystyle \frac{0(05)}{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.

#3
March 6th, 2020,
08:38
Yes, the correct answer is $ \displaystyle \frac{25}{36}$.

MHB Craftsman
#4
March 6th, 2020,
22:32
Thread Author
OK, thanks for the clarifications...