Ten minutes until assignment is due...can anyone help me with this question?
Ok using our basic techniques, do you understand how to find f'(x)?
It is given isnt it?
\(\large\rm f(x)=-3g(x)+5x+5\) \(\large\rm f'(x)=?\)
-3g(x)+5
Hmm you didn't differentiate your g(x)? :O
that is some weird formatting :/ find the derivative of f where is the g(x) anyway?
Why did g(x) stay the same when you took derivative of f(x)?
:P nevermind. through the weird formatting, I found it
according to the info g'(x) @ 5 is 2x-2
whoa whoa whoa y = g(x) y = 2x-2 so 2x-2=g(x)?
zepdrix is right, you need the derivative of g(x) as well because there's product rule going on for f(x) = -3g(x) term
No product rule. You just simply didn't take a derivative of the entire first term.
-3g(x) leave -3 alone deal with g(x) which is g'(x) + leave g(x) deal with -3 which is 0 due to -3 being a constant. unless you're treating g(x) as a constant too.
Im a little lost
Where you at banana man? You simply forgot to take the derivative of the first term. \(\large\rm f(x)=-3g(x)+5x+5\) \(\large\rm f'(x)=-3g'(x)+5\) understand?
yea definitely. back on track.
So now we evaluate this derivative at x=-5, \(\large\rm f'(-5)=-3g'(-5)+5\)
you need g(x) to grab the derivative of g(x) because finding f'(-5) ahahahhahah ... stuckage
If y=2x-2 is the equation of the line tangent to g(x) at x=-5, then g'(-5) gives us THE SLOPE of that equation, ya?
Yup
-12
So which value is the slope in the equation? y=2x-2 m=2, yes? Therefore g'(-5)=2.
-12?? Hmm...
y=g(x) y=2x-2 g(x) = 2x-2 g'(x) = 2
I put 41 but got it wrong
g'(-5) = 2 so now f'(-5) = -3g'(-5)+5 we know that g'(-5) = 2 so plug that in f(-5) = -3(2)+5 now solve that
shoots forgot the ' for f'(-5)
f'(-5) = -3g'(-5)+5 we know that g'(-5) = 2 so plug that into the equation f'(-5) = -3(2)+5 now solve
got it, thanks
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