Question

Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x), where F and G are the functions whose graphs are shown.

http://www.webassign.net/scalcet7/3-2-050.gif

(a) Find P '(2). (b) Find Q'(7).

Answer 1

\[P'(2)=F'(2)G(2)+G'(2)F(2)\] and now we need to find those numbers

Answer 2

\[F(2)=2, G(2)=1\] from right off the graph. \(F'(2)=0\) because the slope of the tangent line there is zero, and \(G'(2)=\frac{1}{2}\) because G is a line with slope \(\frac{1}{2}\)
plug in the numbers to get your answer

Answer 3

thank you so much, could you help with Q'(7). The answer i get isn't right but i am doing it the right way. I find the slope of the line to get F' and G' and than use the quotient rule. however, it's not the right answer

Answer 4

ok let me check again
you are using
\[\frac{g(2)f'(2)-g'(2)f(2)}{g^2(2)}\] right?

Answer 5

nope that was wrong
\[\frac{g(7)f'(7)-g'(7)f(7)}{g^2(7)}\]

Answer 6

yup, i use that and get 3.58 but apparently it is wrong

Answer 7

oh man i totally misled you on the first one

Answer 8

the derivatives were right, but i didn't see that the x-axis was below

Answer 9

no worries i already had the first one i forgot to take it away

Answer 10

i just needed 2

Answer 11

ok

Answer 12

\[f(7)=5, g(7)=1, f'(7)=\frac{1}{4}, g'(7)=-\frac{2}{3}\]

Answer 13

so it should be
\[\frac{1}{4}+\frac{2}{3}\times 5\]

Answer 14

does that work out?

Answer 15

nope that's what i got before and it just says it is wrong, there must be a glitch in the website or something. thank you for your help though

Answer 16

wow ok. sorry it didn't work out

Answer 17

its okay