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OpenStudy (anonymous):
g'(f(x))*f'(x) = g'(f(4)*f'(4)
OpenStudy (anonymous):
f(4) = 8
OpenStudy (anonymous):
so g'(8)*f'(8)?
OpenStudy (anonymous):
so x = 16 and the slope is -1
OpenStudy (anonymous):
so is u'(4) = -1?
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OpenStudy (phi):
to do this problem you need g(x) and f(x)
g(x)= -x + 16
f(x) = 2x for 0≤x<8
g'(x) = -1
f'(x) = 2
g'(8)*f'(4) = -1*2 = -2
notice that because g' and f' are constant, the x values are irrelevant for this problem.
for u(12) we must use a different function for f(x)
OpenStudy (anonymous):
okay so then u'(12) = 1 correct?
OpenStudy (anonymous):
i mean -1 , sorry
OpenStudy (anonymous):
wait no it must be 2 !!
OpenStudy (anonymous):
i get it now :) ty guys!!!
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OpenStudy (phi):
what are you using for f(x) when x>8 ?
it should be
f(x)= -2x + 32
notice f(8)= 16, and f(16) =0, as it should
g(x) =-x + 16
as a check, we can form
u(x)= g(f(x)) = -( -2x + 32)+16 = 2x -16
and u' = 2
(independent of x)
using the chain rule, you should get the same answer.