1)
 = \frac{f(1+5x)+f(1+6x)}{x})
Given
f(1) = 0, and
f '(1) = 8, evaluate
)
Limits as it results to
, so differentiate top and bottom using Chain Rule:
f'(1+5x)+(6)f'(1+6x)}{1} = 40+48 = 98)
 \mbox{ is continuous at } x = 1,})
2) If
f(2) = 13 and
f '(
x) ≥ 1 for 4 ≤
x ≤ 4, how small can
f(4) possibly be?
f(4) - f(2) = f'(c)(4-2)
f(4) = 13 + (2)f'(c) > 1
?
The last step, and where to go from there is what I'm a little shaky on.
=13+2f'(c)>13+2=15})
3)
^q)
where q = hx
h and c are both constants, to my knowledge.
We were taught to do this in a way that seems a little convoluted, but here's as far as I got with this one:
=
^q = e^g)
^q=\left(\lim_{x\rightarrow\infty}\le ft(1+\frac{c}{x}\right)^x\right)^h=e^{ch}})
; g = (hx)ln(1+\frac{c}{x}))
From that point we can just take the limit of e to the g'(x), so
 = (hx)ln(1+\frac{c}{x}))
=
}{\frac{1}{hx}})
Then I get stuck again, because the derivative of the bottom is
^2})
I'm imaging there has to be a way to simplify that, but I can't think of it.
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Originally Posted by Open that Hampster!
1)
