Question

For all reals numbers x and h, let f be a continuous function such that

i) (f(x + h) = f(x) + f(h) + 2xh for all x and h

ii) lim as h approaches 0 of f(h) / h = 7

a) Find f(0). Justify

b) Use the definition of the derivative to find f'(x). Show your work.

http://imgur.com/a/ngOt2

Answer 1

@phi is my part b correct? http://imgur.com/a/3dJ0i

Answer 2

the definition of f'(x) is
\[ \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} \]
use that a your starting point.

Answer 3

That's what I did

Answer 4

Just with f'(x + h) instead of f'(x)

Answer 5

no. we don't do f'(x+h)
we use the definition
\[ f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} \]
the first line should start like that
next, replace f(x+h) in that mess
then simplify
what do you get ?

Answer 6

@phi I get f(x) + f(h) + 2xh - ((f(x+h) - f(h) - 2xh)) / h.
Simplify this to get (f(x) - f(x+h) + 2f(h) + 4xh) / h

Am I then comparing with the limit in part ii?

Answer 7

almost. but the 2nd part is *just* f(x)
you are doing f(x+h) - f(x) NOT f(x+h)-f(x+h)

Answer 8

in other words, you only replace f(x+h) with f(x)+f(h) +2xh in the definition of f'(x)
everything else stays

Answer 9

Ohhh ok

Answer 10

So then it simplifies to (f(h) + 2xh) / h

Answer 11

don't leave out the limit
you get
\[ \lim_{h \rightarrow 0} \frac{f(h)+2xh}{h} \\
\lim_{h \rightarrow 0} \frac{f(h)}{h}+ \lim_{h \rightarrow 0} \frac{2xh}{h} \]

Answer 12

Nice. Thank you so much. Then I just substitute in 7

Answer 13

and of course 2xh/h simplifies to 2x
and the limit is 2x (h has no effect on 2x)
the final answer should be
7+2x

Answer 14

Ah right. Thanks for that.