/sci/ - Science & Math

your pic is the definition of derivative as a limit, the answer is h'(2)

It's like we _don't_ have jsmath here

<div class="math">\lim_{t\to 2} \frac{h(t) - h(2)}{t - 2}</div>

<div class="math">h(2) = 7</div>

No the accepted answer is 3 and the preface to the question is: evaluate:


3 is answer in back of book.,

>>1847802


I don't know how to use it.

look under the form.

>Use TeX/jsMath with the <span class="math"> (inline) and <div class="math"> (block) tags. Double-click equations to view the source.

TeX itself is pretty easy for equations[/spoiler]</div>

>>1847804
>>1847804
There's not enough information to arrive at that.

Ok thanks, any input on how to evaluate this to arrive at 3?

>>1847812


Then I'm so fucking lost. I don't know how the book has a concrete answer.

Is it me or you should be given h(t)?, there are infinite functions that pass through (2,7) and they need not have the same derivative at all

>>1847820


No formula, only that h(2) = 7, and it says that h'(2) = 3, but it still says evaluate the problem. I mean maybe I'm wrong but is this entire series of 5 questions as simple as writing the answer they already gave you? I mean what's the point of a problem presented? So I just write h'(2) = 3, and that's it?

Something tells me if I do that on my test I'm fucked.

I dunno man, but it is clear that you arent given enough info, maybe youre supposed to use the the solution from a previous problem or something?

hint:
(1) h(t) = t^2 - t + 5
(2) h'(t) = 2t - 1

proof
(1) let h(2) = t^2 - t + 5 = 2^2 - 2 + 5 = 4 - 2 + 5 = 2 + 5 = 7

(2) let h'(2) = h(2) dt = 2t - 1 = 2(2) - 1 = 4 - 1 = 3

the rest should be easy.

>>1847827
>it says that h'(2) = 3
Well there you go.

On a test I would write something like
This expression is the definition of the derivative evaluated at x = 2. Since h'(2) is known to equal 3, this expression must, too.

>>1847844

given that h(t) = t^2 - t + 5 and h(2) = 7

lim t>2 [ (h(t) - h(2)) / (t - 2) ]
> lim t>2 [ (h(t) - 7) / (t - 2) ]
> lim t>2 [ ((t^2 - t + 5) - 7) / (t - 2) ]
> lim t>2 [ (t^2 - t + 5 - 7) / (t - 2) ]
> lim t>2 [ (t^2 - t - 2) / (t - 2) ]
> lim t>2 [ ((t+1)(t - 2)) / (t - 2) ]
> lim t>2 [ (t+1) ]
> [ 2+1 ]
> 3

*bing*

>>1847844
>>1847883

i hereby proclaim myself to be the best mathematician currently lurking /sci/

>>1847844
Lol no. Not enough information to give such an equation.

>>1847897
>>1847883

give me a fucking break.
you said h'(2) = 3
this is simple shit here, son! standard solution.
i don't have your manual. i just pulled this out of my hat. this is how you think like a mathematician, bitch.

>>1847897
listen: since this is a problem with derivatives, you may as well assume that the function is at least quadratic, otherwise it's derivative will be a real number and therefore it can't possibly be the answer.

since h(2) is known to equal 7 then it is necessary to construct a function which, when computed, equals 7.
since this function must be quadratic then it is simple to arrive at such a function which would produce the necessary answer, in this case, 7.

h(t) therefore must be t^2 - t + 5 (it could possibly be another complex solution but this one is the easiest).

running the first equation through with (t^2 - t + 5) in place of h(t) and 7 in place of h(2)

computing this produces the desired answer.
as you can see, nothing more than h(2) = 2 and the original information is necessary to produce a detailed solution to the problem.

ps i'm very bored and sort of ignoring my responsibilities at the moment but i hope you take this lesson to heart and learn to understand calculus in a holistic way.

captcha: Königin heinate

>>1848012
>>nothing more than h(2) = 2
h(2) = 7
srry