Calculus homework

s.h_math Author

wrote...

#14

2 years ago Edited: 2 years ago, s.h_math

do you mean where it says -f(x) that's the second neg that makes it positive?

Post Merge: 2 years ago

do you mean where it says -f(x) that's the second neg that makes it positive?

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bio_man

wrote...

#15

Educator

2 years ago

Quote from: s.h_math (2 years ago)

do you mean where it says -f(x) that's the second neg that makes it positive?

Exactly 👍

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s.h_math Author

wrote...

#16

2 years ago

oh okay I believe I understand. Thank you. basically, if it was a positive 5, in the equation I would sub in -5, correct?

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bio_man

wrote...

#17

Educator

2 years ago

Technically, yes. We got -5 as the output, then we put it as f(x) in the formula which had a negative at the front, for example

... - f(x)
... - (-5)
... + 5

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s.h_math Author

wrote...

#18

2 years ago

also sir one last question, where do you get (h-2) from?

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bio_man

wrote...

#19

Educator

2 years ago

You see where we simplified to h^2 - 2h? I common factored an 'h' from it, so:

h(h-2)

Then, given that the denominator had an 'h' also, it cancelled with the factored 'h' above, leaving me with h-2 as my final expression.

Do you see it?

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s.h_math Author

wrote...

#20

2 years ago

wouldn't it be done like this sir?

Post Merge: 2 years ago

https://ibb.co/RbBndHB

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bio_man

wrote...

#21

Educator

2 years ago

Quote from: s.h_math (2 years ago)

wouldn't it be done like this sir?

Post Merge: 2 years ago

https://ibb.co/RbBndHB

You could do it that way where you don't evaluate f(2) until the very end, as you did in your sample.

What I did was evaluate x right from the beginning -- doesn't matter either way.

Also, I didn't take the limit as h->0. But if I did, I would get the same answer:

\(m = h - 2\)

\(m\ =\ \lim _{h\ \rightarrow 0}\left(h-2\right)\)

\(m\ =\ -2\)

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s.h_math Author

wrote...

#22

2 years ago

Thank you so much for all your help!

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