quotient rule: f(x)= (8x-4) / (2+2x) ; x=2

[yanarains]

f(x)= (8x-4) / (2+2x) ; x=2

=2x(8x-4) - (2+2x)x8 / (2+2x)^2
=16x-8 - 16 -16x / (2+2x)^2

I then substitute x for 2

-24/36 = -.667 (direction are to round to the 3rd decimal place)
correct answer is .667????????? Why is this a positive answer and not negative? I must be missing the final step. What is the next to step to figure if the answer i s positive of negative.

Thanks for your help.

 

[galactus]

I think you may have a few little mistankes.

\(\displaystyle \L\\\frac{8x-4}{2+2x}\)

\(\displaystyle \L\\\frac{(2+2x)(8)-(8x-4)(2)}{(2+2x)^{2}}\)

 

[stapel]

f(x)= (8x-4) / (2+2x) ; x=2

I will guess that the instructions were something along the lines of "Use the Quotient Rule to find the derivative of f(x), and then evaluate f'(2)." If this guess is incorrect, please reply with clarification.

=2x(8x-4) - (2+2x)x8 / (2+2x)^2
=16x-8 - 16 -16x / (2+2x)^2

I will guess that you meant to have an f'(x) in front of the first "equals" sign, that you used "x" to mean both "the variable" and "times", and that you forgot some grouping symbols, so you actually meant the following:

. . . . .f'(x) = [(2x)(8x - 4) - (2 + 2x)(8)] / [2 + 2x]^2

. . . . . .. . .= [16x - 8x - 16 - 16x] / [2 + 2x]^2

. . . . . .. . .= [-8x - 16] / [2 + 2x]^2

However, the Quotient Rule is not as you have stated. :shock:

Rather, the Rule is as follows:

. . . . .For f(x) = g(x)/h(x), for h(x) non-zero, f'(x) is given by:

. . . . . . . . .f'(x) = [g'(x) h(x) - g(x) h'(x)] / [h(x)]^2

That is, you reversed the subtraction in your numerator. Correct that (and correct your notation), and you should get the right answer!

Eliz.