Desperate Help Needed for Calculus.

[junior bluff]

Ask chatgpt or something

[safe bay]

@dry temple
Choose arbitrary point $k \in (c_1, c_2)$ s.t. $f(k) < f(c_1) = f(c_2)$ \
Let $\frac{f(k) - f(c_1)}{k - c_1} = p$. $\because p < 0$, $\therefore$ By MVT, $\exists d \in [c_1, k]: f'(d) = p$ \
Similarly, let $\frac{f(k) - f(c_2)}{k - c_2} = q$. Let $e \in [k, c_2]: f'(e) = q >0 $. \
By IVT, $\exists p \in (d, e): f'(p) = 0$

[tender phoenix]

Kelvin Chan (Tag me 2 reply)

[safe bay]

@junior bluff That's the most meaningful mathematical answer that I have ever seen in my life.

[dry temple]

Is this all that is needed?

[junior bluff]

Thank you, I am truly flattered.

[thorn bough]

@junior bluff has given 1 rep to @safe bay

[safe bay]

...Yes...You only need need MVT and IVT.

I was thinking of using Nested Interval Property, but it seems that there is more simple answer.

[dry temple]

What you have written is suffice to prove the given statement right?

[dry temple]

How have you shown that f has a local minima on (a,b) tho?

There was no mention of a or b in what u have written.

Please go to the general chat.

[junior bluff]

Alright

[safe bay]

Yes, you are right...let me think again. Sorry for it

[dry temple]

You have to use MVT, IVT or EVT. You just didn't prove what was needed lol.

Lemme know if u get it.

[safe bay]

$f$ has a local min. in $[c_1, c_2]$ implies that $f$ has a local min. in $[a, b]$.
My only problem is just that $d$ might be a local max.

[tender phoenix]

Kelvin Chan (Tag me 2 reply)

[west kettle]

The way I see it is something like this

Horribly drawn cuz rn I'm using phone but the idea is

If there are 2 local maxima

That means it needs to go down somewhere in between c1 and c2

Otherwise one of them won't be a local maxima anymore

[west kettle]

No matter how u draw it, u won't be able to make 2 peaks in the graph without first going down after drawing the first peak, creating a local minimum

[safe bay]

I am thinking whether I can use f([c_1, c_2]) is a closed interval, say f([c_1, c_2]) = [c, d], and claim that e s.t. f(e) = d is a local minimum.

$\forall x \in (c, c_2): \frac{f(x) - f(c)}{x - c} > 0$, $\therefore f'(e^+) \geq 0$
Similarly, $f'(e^-) \leq 0$.
$\therefore f'(e) = 0$

[tender phoenix]

Kelvin Chan (Tag me 2 reply)

[safe bay]

The assumption that $c_1$ and $c_2$ are local max. is important here since $e$ would not be one of $c_1$ or $c_2$.