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well their is:
a)13/2
b)7/2
c)-5/2
d)11/2
e)17/2

So what did you get for f'?

I will check it.

\[f'(x) = h(x) + -2h'(x)+ (\frac{ 2h(x)-(2x)h(x) }{ h(x)^2 }\]

Also the answer is there.
h(x) shouldn't be there.
you are missing h' in your quotient

(-2h)'=-2(h)'=-2h'

oh ok

(2x/h)'=[(2x)'h-2x(h)']/h^2

so the awnser should be 19/2

hmm... that isn't what I got...let me check my result.

Oh you are still assuming h is there aren't you?

(-2h)'=-2(h)'=-2h' not h-2h'

So how does it apply

what do you mean?

correct?

Well see I keep telling you (-2h)'=-2h' not h-2h'

That first term you are putting in shouldn't be there.

Could you show me what you plugged in so I understand

So you don't understand why (-2h)' is -2h' and not h-2h'?

No

I'm liking I'm using the constant multiple rule.

The constant multiple rule:
k=cf
k'=cf'

So when I plug in for -2h' when h(4)=4 and h'(4)=-2

Yep

-2h'(x)
x is 4.
-2h'(4)
-2h'(4)=-2(-2)=4
Now simplify your quotient and add.

Does this mean -2h'= -2(-2) or -2(4)

ok

h' is h'(x)

This is what you wrote earlier
I'm telling you (-2h)'=-2h' not h-2h'
do you understand now?

Oh ok Now I get what you mean by this

I put the marks through the bad parts.

so you understand the constant multiple rule?

Yes I do now understand the multiple rule and the awnser is 11/2

That is right.

I just didnt understand it conceptually but now that I see it plugged in and used I do now

Ok. Neatness.

Thank you so much

Np.