anonymous
  • anonymous
Q:f={(0,5),(1,4),(2,7),(3,14)} and g(x)=3f(x+1)+7 a)evalute g(-1) b) evalute g (1) c) evalute g(3)
Mathematics
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anonymous
  • anonymous
Q:f={(0,5),(1,4),(2,7),(3,14)} and g(x)=3f(x+1)+7 a)evalute g(-1) b) evalute g (1) c) evalute g(3)
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
a)22 b)28 c)is notdefined
anonymous
  • anonymous
i noe but how did u solve it can u show me tht
anonymous
  • anonymous
sure

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anonymous
  • anonymous
a) g(-1)=3f(-1+1)+7=3f(0)+7 from the definition of f we see that f(0)=5 so g(-1)=3f(0)+7=3(5)+7=22
anonymous
  • anonymous
g(x) = 3f(-1 + 1) + 7 g(x) = 3f(0) + 7 find f(0) from the Q:F (0,5) g(x) = 3*5 + 7 g(x) = 15 + 7 g(x) = 22 g(x) = 3f(1 + 1) + 7 g(x) = 3f(2) + 7 find f(2) = (2,7) g(x) = 3*7 + 7 g(x) = 21 + 7 g(x) = 28 g(x) = 3f(3 + 1) + 7 g(x) = 3f(4) + 7 find f(4) = not listed don't know what to do
anonymous
  • anonymous
f is explicitly defined as the given set of ordered pairs, therefore there is nothing to "do" in part (c) , it is simply not defined
anonymous
  • anonymous
thannks :)

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