The graphs of the function F (left, in blue) and G (right, in red) are below. Let P(x)=F(x)G(x) and Q(x)=F(x)/G(x). Answer the following questions. P'(1), Q'(1), P'(6), Q'(6)
the graphs: http://roy.math.umn.edu/webwork2_course_files/umtymp-calculus1-f13/tmp/gif/muzzammil-Q-raza-617-setDerivative_Rulesprob22image1.png http://roy.math.umn.edu/webwork2_course_files/umtymp-calculus1-f13/tmp/gif/muzzammil-Q-raza-617-setDerivative_Rulesprob22image2.png
the first if F and the second is G
@dumbcow @wio
@zepdrix
@ash2326 @hartnn @Itsjustlaciee @thomaster
Use product rule and quotient rule to compute \(P'(x)\) and \(Q'(x)\). To evaluate this at x=1,6 you need to read of F'(1),F'(6),G'(1),G'(6) from the given graph
how do i find F'(1) and etc with only the graphs
what F'(1) means geometrically?
not sure
is there any connection between derivative and tangent line?
derivative is slope of tangent line
correct and lucky for you the graphs of F,G at points x=1, x=6 are straight lines so derivative is simply the slope of the line
ok. so i find the slope at those points and then do what
but is it a straight line for F
input those values into the product rule/quotient rule for F'(x), G'(x)
oh right F(1) is vertex of parabola, well imagine drawing a tangent line through that point....what is its slope?
0
yep
ok but how do i find F'(x) and G'(x)
i dont know their equations
F'(x), G'(x) are just the slopes so we found slope at F(1) is 0 right...so F'(1) = 0
ok and G'(1)=3
no look at slope of the line again
rise/run=3/1
are you looking at line to right? when x=6 the line has slope of 3 G'(6) = 3
oh sorry yeah its 1. i am doing both parts at the same time
ok then i get F'(6) = 1/4
ok
would G'(1)=1?
yes you had it right
ok thanks. i know how to get it now
never mind i did something wrong
F'(1)=0 F(1)=1 G'(1)=1 G(1)=2 F'(6)=1/4 F(6)=3 G'(6)=3 G(6)=3
@dumbcow
this is what i got P'(1)=1 Q'(1)=1/4 P'(6)=39/4) Q'(6)=13/12
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