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Calc question

Anonymous Mon 15 Jun 01:57:43 2020 No.11799393 View Reply Original Report

Quoted By: >>11799937 >>11800006 >>11800028 >>11800071

What am i supposed to plug in? i thought it would be (2(1+h)^2-7)/h then plug in 0 but that doesn't work.

Anonymous

Anonymous Mon 15 Jun 2020 05:30:38 No.11799937 Report

Quoted By:

>>11799393
>i thought it would be (2(1+h)^2-7)/h
How did you get this result? Do you understand what notation f(a) means?
Even disregarding the fact that this expression is wrong, why is that
> that doesn't work.
?
You can calculate this limit (yes, even by 'plugging in 0') and get a non-zero result (but again, it is not the answer to the question as the expression is wrong in the first place).
This can be solved either manually how you are trying to do it (and this is trivial and you should be able to do it effortlessly, so be patient), or by recognising that this is an expression for something else.

Anonymous

Anonymous Mon 15 Jun 2020 06:02:38 No.11800006 Report

Quoted By:

>>11799393
You're evaluating a limit for h
2(1+h)^2 +5 - (2*1^2 +5) all over h
gets you (4h+h^2)/h after expanding (1+h)^2 and resolving f(1) to a numerical value

Anonymous

Anonymous Mon 15 Jun 2020 06:10:10 No.11800028 Report

Quoted By:

>>11799393
4x then plug in 1
4
https://www.google.com/search?q=y%3D+2x%5E2%2B5%2C4x

Anonymous

Anonymous Mon 15 Jun 2020 06:24:27 No.11800064 Report

Quoted By:

I'm not going to tell you the answer because this is basic stuff that you need to understand how to do but I will tell you that this operation is well behaved for any polynomial you use it on and will be defined for any input x, so DNE will never be the answer for a polynomial.

Anonymous

Anonymous Mon 15 Jun 2020 06:28:42 No.11800071 Report

Quoted By:

>>11799393
$ f(x)=2x^2+5 \\ \displaystyle \lim_{h \to 0} \dfrac{f(1+h)-f(1)}{h} = \lim_{h \to 0} \dfrac{2(1+h)^2+5 -(2\cdot 1^2+5)}{h} \\ \displaystyle = \lim_{h \to 0} \dfrac{2(1^2+2 \cdot 1 \cdot h+h^2)+5 -(2 \cdot 1^2+5)}{h} = \lim_{h \to 0} \dfrac{4h+h^2}{h} = \lim_{h \to 0}4+h = 4 $

Anonymous

Anonymous Mon 15 Jun 2020 06:31:33 No.11800080 Report

Quoted By:

$ f(x)=2x^2+5 \\ \displaystyle \lim_{h \to 0} \dfrac{f(1+h)-f(1)}{h} = \lim_{h \to 0} \dfrac{2(1+h)^2+5 -(2\cdot 1^2+5)}{h} \\ \displaystyle = \lim_{h \to 0} \dfrac{2(1^2+2 \cdot 1 \cdot h+h^2)+5 -(2 \cdot 1^2+5)}{h} = \lim_{h \to 0} \dfrac{4h+2h^2}{h} = \lim_{h \to 0}4+2h = 4 $

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