Posted by catherine on Tuesday, September 14, 2010 at 10:02pm.
please show your work. maybe myself or the other tutors could figure out what you did wrong.
ok. this question also had a part a) which was
(a) What is f '(3)?
so i just divided
(9h^2+8h)/h
(h(9h+8))/h and plugged in 0 since lim->0
got 8 as my answer.
now for B)
i tried plugging 3 into (h) in func 9h^2+8h/h
did same with 7 and got wrong answer. i tried many different ways but kept getting wrong answer.
what i am confused is with how to find values for f(7) and f(3) since there is no function given to begin with...
There is no limitation on the value/size of h, so h can be any number.
f (3 + h) - f (3) = 9h^2 + 8h
so
f(3+h) = 9h^2 + 8h - f (3)
Try setting h=4 to see what you get.
Sorry, the equation should read:
f(3+h) = 9h^2 + 8h + f (3)
In fact, the solution is simpler than that.
Remember TutorCat said:
for the secant:
[f(7)-f(3)]/(7-3)
you can work out
[f(7)-f(3)]
from
f (3 + h) - f (3) = 9h^2 + 8h
by putting h=?
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